'm^^ 


y    -*?'^- 


v(:\\" 


PE AOTICAL  ANT»    *'Tr,  OliETTCAL 


A  E  I  T  H  M  E  T  I  C , 


IX   ADBITION    TO   TUS   USUAL   ilOPP^  OF   ^  PKRATTON,    THE   SCI- 
OF   NUMBEllS,    THE   PKCh',  'x   SYSTEM, 

AND   OTHER   TilTOi:  ?, 

HCLP   A  p:; 


BY  HORATIO  N.  ROBiNSOxV,  A    M., 

AUTHOR    OF     V    C;)l;u-[;    Of    >rATHE:M.>.av::3. 


JOUS  COLLEGES,  AXD  PKIYATJl  STUDFXTS. 


JACOB 


[N  STREET. 


RY 

Y  Of      I 

NIA       I 


f  i?SITY  Of 
LIK>RH1A 


«X(HIUnB. 


^^ 


•    •    • 


d?>:p 


ROBINSON'S 

MATHEMATICAL, 

PHILOSOPHICAL,  AND  ASTRONOMICAL 

CLASS  BOOKS, 

PUBLISHED  AND  FOR  SALE  BY 

JACOB     EHNST, 

lis  JUalBi  street,  €iBiciEBii:^ti. 


FOR    SALE    IN 
boston: 

B.  B.  VU6SET  ft  CO. ;    E.  DAVIS  !c  CO. ;    W.  J.  REYNOLDS  b  CO. ;    PHILIJPS.  SAMPSON  b  CO. 

NEW    YORK: 

MASON  BROTHERS ;   D.  BURGESS  &  CO. ;   NEWMAN  &  IVISON  ;  PRATT,  VTOODFORD  ft  CO ; 
A.    S.   BARNES  ft   CO. 

PHILADELPHIA! 

UPPINCOTT,   QRAMBO    ft   CO.;    THOMAS,   COWPERTHVTAITE  ft  CO.; 
E.  H.  BUTLER  k  CO.;   URIAH  HUNT  &  SON. 

BUFFALO,     N.  Y. 

FHINNET  &   CO. ;    DERBY,    ORTON    &  MULUGAH. 

SYRACUSE,     N.  Y. 

E.   U.   BAECOCK   k  CO. 
AND   THE   PRINCIPAL   BOOKSELLERS   IN   THE   UNITED   STATES. 


A 


A    NEW 

PKACTICAL  AND  TIIEOllETIC AL 
ARITHMETIC, 

^       IN    WHICH, 


IN  ADDITION  TO  THE   USUAL  MODES  OF  OPERATION,  THE   SCIENCE 

OF  NUMBERS,  THE   PRUSSIAN  CANCELING  SYSTEM,  AND 

OTHER  IMPORTANT  ABBREVIATIONS,  HOLD 

A  PROMINENT  PLACE. 


BY  HORATIO  N.  ROBINSON,  A.  M. 

AUTHOR    OF    A    COURSE    OF    MATHP:mATICS. 


DESIGNED  FOR  SCHOOLS,  COLLEGES,  AND  PRIVATE  STUDENTS.       \ 

JOHM  S.  PV^ELL 

Civil  (  .         ^Ineer, 

8AN  FRANCl^Cg,  GaL 

CINCINNATI* 
JACOB  ERNST,  112  MAIN  STREET 

1854. 


woaaia  ua. 


Entered  according  to  Act  of  Congress,  in  the  year  1845, 

BY  E.  MORGAN  &  CO., 

In  the  Clerk's  olfice  for  the  District  Court  of  Ohio. 


^U^Ck  1 


Stereotyped  by  J.  A.  Jamee, 
Cincinnati. 


h 

ED'JC. 
UBRARY 


Civil  &  Mechanical  Engineer. 

SAN  FRANCISCO,  CAL. 
PREFACE. 


The  public  may  very  properly  inquire,  what  <roocl  can  be 
accomplished  by  addino;  another  Arithmetic  to  tlie  long  list 
now  in  existence]  Both  teachers  and  the  taufjht  are  already 
sufficiently  perplexed,  in  making  a  selection. 

True,  there  are  many  arithmetics  before  us,  claiming  our 
attention;  but  more  than  half  of  them  are  mere  collections  of 
questions,  ordered  to  be  solved  by  rules,  given  in  a  spiritless 
manner,  having  no  connected  system,  and  not  recognizing  the 
general  and  universal  scientific  character  of  numbers. 

They  place  arithmetic  before  the  mind  of  the  pupil,  as  an 
art,  not  as  a  science — they  give  him  directions  ichat  to  do,  not 
what  to  think,  and  how  to  reason;  and,  in  short,  many  of 
these  books,  now  pressing  for  patronage,  are  already  con- 
demned by  all  good  judges  and  scientific  teachers. 

Within  the  last  ten  or  fifteen  years,  the  science  of  arith- 
metic has  undergone  many  changes,  and  received  many  im- 
portant improvements ;  and  such  improvements  have  appeared, 
little  by  little,  and  from  time  to  time — some  in  one  book,  and 
some  in  another — nowhere  presenting  one  grand  whole;  and 
the  object  of  this  work  is,  to  give  unity  and  system  to  all  the 
modern  improvements  \\\\\c\\  preseyii practical  utility^  and  in- 
terweave them  in  the  common  and  general  system,  or  such 
parts  of  it  as  are  necessarily  retained;  and,  how  far  we  have 
accomplished  this  object,  let  the  competent,  the  unprejudiced, 
and  the  uninterested^  answer. 

These  modern  improvements  generally  pass  under  the  name 
of  the  canceling  system  ;  but  canceling  by  no  means  includes 
them  all,  nor  are  they  all  included  in  numerical  operations- 
principles,  explanations,  and  forms,  have  been  simplified  and 
improved. 

Durin-g  the  progress  of  investigation,  there  needs  must  be 
many  fruitless  attempts  at  improvement,  and  much  useless 
matter  must  accumulate  on  the  hands  of  original  inquirers; 
and  some  such  matter  occasionally  appears  to  the  vv  orld,  rath- 
er because  it  is  singular,  or  new,  than  because  it  is  practical 

010 


VI  PREFACE. 

and  useful;  but  we  have  carefully  avoided  all  that  we  did  not 
deem  of  practical  utilit}^,  either  in  teaching  principles,  or  in 
facilitating  numerical  computations.  Several  small  books 
have  appeared  (evidently  not  written  by  men  of  science),  pro- 
fessing- to  teach  arithmetic  in  a  few  lessons,  on  modern  im- 
proved plans,  &c.;  but  they  are  mere  outlines  of  rules  and 
forms,  unconnected  with  any  general  system  ;  and,  of  course, 
they  must  fail  to  teach  arithmetic  as  a  whole,  or  long  hold 
public  attention. 

Some  other  authors,  more  fond  of  quantity  than  of  quality, 
not  discriminating  between  the  singular  and  the  useful,  have 
collected  together,  in  one  great  mass,  a//that  is  new,  and  giv- 
en it  to  the  world  as  improvements,  and  claim  merit  and  pat- 
ronage, because  they  have  given  all.  As  well  might  the  far- 
mer claim  merit  and  compensation,  for  bringing  straw  and 
chaff  to  market  with  his  wheat. 

To  make  ourselves  thoroughly  understood  on  this  point, 
we  remark  that  we  have  given  many  useful  and  practical 
methods  of  multiplication;  but  few  or  no  corresponding  meth- 
ods of  division,  as  the  corresponding  methods  of  division,  we 
found,  would  not  be  practically  useful.  For  example,  there 
are  many  practical,  obvious,  and  easy  methods  of  multiplying 
any  niLmber :  say  12468  by  9,  99,  98,  97,  or  any  number  near 
10,  100,  1000,  &c.  And  there  are  corresponding  methods  of 
dividing  any  number  by  9,  99,  98,  &c.;  but  the  operations  are 
not  obvious  and  easy,  as  in  multiplication,  we  do  not  regard 
them  as  improvements;  for  the  common  way  is  full  as  easy, 
and  we  will  not  tax  the  temper  and  patience  of  the  pupil,  by 
taking  him  off  into  speculations  of  questionable  utility. 

For  those  fond  of  speculation,  however,  we  will  give  the 
method  of  dividing  by  9,  or  99,  98,  &c.;  and  those  who  please, 
may  extend  the  principle ;  but  it  is  rather  too  speculative  for 
the  common  pupil.  Divide  1836  by  9.  Divide  by  10,  and 
the  quotient  will  be  too  small,  because  the  divisor  is  then  too 
large;  but  divide  the  quotient  by  10,  continually,  as  far  as 
tenths:  thus. 

Divide  by  10, 183,  6 

Again,  by  10, 18,  3 

Again,  by  10, 1,8 

Again,  by  10, 1 

Add,  and  we  have,     .    .     Quo//en^,  202,18,  remainder. 

This  remainder,  18,  will  divide  by  9,  giving  2  to  be  added 

to  202,  making  204  for  the  actual  quotient;  but  how  much 

I  more  readily  1836  could  have  been  divided  by  9,  in  the  usual 


PREFACE. 


way  1     There  is  now  and  then  a  case  where  this  method 
might  be  preferable.     This  is  one  :     Divide  3436  by  99. 

"^  Divide  by        100, 34,  3G 

Divide  this  by  100, 34 

Quotient,  34,  70,  Remainder. 
Here  is  a  more  difficult  case.     Divide  49999  by  98. 

Quo.   Re. 

Divide  by  100, 499,  99 

Double  this  quotient,  and  divide  by  100,  gives     9,  98 
Double  9,  and  divide  by  100, 18 

Add  up,  as  in  decimals.  Quotient,     510,   15 

As  the  divisor  was  98,  each  100  must  have  a  remainder  of 
2,  and  the  remainders  summed  up  to  more  than  200,  our  re- 
mainder, 15,  must  be  increased  by  4 ;  then  the  quotient  is 
510 — remainder  19.  This  may  be  curious,  but  the  common 
way  is  much  better. 

This  may  be  new,  but  it  is  not  improvement,  and  by  throw- 
ing all  things  together,  without  discrimination,  some  have 
brought  real  excellence  into  disrepute;  and,  therefore,  we 
have"  taken  much  care  not  to  introduce  any  matter  that  we 
could  not,  for  real  utility,  recommend.  Clearness  and  per- 
spicuity have  been  our  constant  aim  ;  and,  although  we  set  a 
very  high  price  on  brevity,  it  has  never  been  purchased  at  the 
expense  of  perspicuity. 

The  reader  will  observe,  that  we  have  laid  great  stress  on 
the  modern  improvements.  True,  we  have  so;  but  there  are 
really  no  new  principles — there  may  be  new  combinations  of 
them — there  are  no  new  mechanical  powers,  but  there  are 
new  machines  every  day;  and  it  is  not  too  much  to  say,  that 
no  arithmetician,  however  great  his  knowledge,  can  be  con- 
sidered very  skillful,  unless  he  has  paid  particular  attention 
to  this  branch  of  study  within  a  very  few  years.  Yet,  in 
many  respects,  the  common  and  ancient  modes  of  operation 
must  ever  be  employed  ;  and,  in  truth,  are  the  best.  It  must 
not  be  understood  that  we  have  abandoned  them,  because  we 
have  brought  improvements  to  them. 

For  peculiarity  and  brevity  in  numerical  operation,  the 
reader  is  referred  to  the  "  Supplement  to  Multiplication  and 
Division,  Reduction,  Proportion,  Interest,  Exchange,  Mensu- 
ration, and  the  roots." 

To  the  method  of  stating  problems  in  proportion,  by  com- 
paring cause  and  effect^  we  invite  special  attention.  On  criti- 
cal examination,  it  will  be  found  more  easy  and  more  ration- 
al than  any  other  method. 


PREFACE. 


Other  methods  of  statement  sometimes  require  the  number 
of  men  to  be  multiplied  into  feet  of  wall — days  into  acres  of 
grass,  &c. — all  of  which,  though  correct  as  abstract  propor- 
tion in  numbers^  is  unnatural  and  void  of  strict  philosophical 
expression ;  not  so  with  this  method. 

The  peculiarities  which  the  student  will  here  find  in  the 
extraction  of  roots,  and  in  mensuration,  are  not  all  new — in- 
deed, there  can  be  nothing-  new  in  principlt — but,  as  far  as  the 
author's  knowledg-e  extends,  he  is  not  aware  that  these  abbre- 
viations have  ever  been  collected  in  any  arithmetical  work, 
except  in  the  Appendix  to  Talbott's  Arithmetic,  and  that  very 
recently. 

We  place  much  importance  on  the  canceling  or  factor  sys- 
tem, in  performing  operations  requiring  multiplications  and 
divisions,  for  their  great  utility  in  algebra,  when  the  pupil 
arrives  at  that  science  ;  and  that  teacher  who  looks  no  fur- 
ther than  to  solve  a  few  problems  in  arithmetic,  has  not  a 
sufficiently  extended  view  to  judge  of  the  merits  of  the  sys- 
tem ;  therefore,  let  none  such  condemn. 

In  preparing  this  treatise,  care  has  been  taken  to  avoid  all 
puzzles,  as  being  unworthy  of  science;  and,  all  unusual 
or  difficult  problems  :  the  object  has  been  to  teach  the  science 
of  numbers — not  to  try  the  powers  of  arithmeticians.  We 
deem  it  no  recommendation  to  any  author,  that  his  questions 
are  very  simple,  or  very  severe  ;  we  dislike  extremes,  on  eith- 
er hand.  In  choice  of  problems,  we  have  aimed  at  sound 
practical  utility,  bringing  them  as  near  to  the  real  business 
of  active  life  as  possible.  To  carry  out  these  views,  we 
have  added  a  short  Appendix,  containing  problems  under  the 
mechanical  powers — though  this  is  not,  technically,  arith- 
metic. A  brief  practical  system  of  book-keeping  is  subjoin- 
ed, sufficient  for  the  farmer  and  mechanic,  and  for  all  tliose 
whose  operations  do  not  require  the  more  complicated  system 
of  double  entry. 

In  conclusion,  we  would  remind  the  reader,  that  we  do  not 
claim  to  have  presented  a  perfect  work ;  perfection  is,  per- 
haps, impossible  ;  but  it  has  been  our  design  to  present  a  con- 
nected, simple,  and  scientific  system,  sufficient  for  the  busi- 
ness man  and  scholar,  without  being  encumbered  witii  unne- 
cessary theories;  yet,  uniting  and  systematizing  all  real  im- 
provements, taking  only  the  practical  and  useful;  and,  how 
far  we  have  attained  these  objects,  an  impartial  public,  aided 
by  experience,  must  determine. 


CONTENTS 


Numeration, Page  12 

Simple  Addition, 16 

Simple  Addition,  applied  to  Federal  Money, ....  20 

Simple  Subtraction, 21 

Subtraction  in  Federal  Money, 23 

Simple  Multiplication, 24 

Simple  Division, 32 

SECTION  II. 

Supplement  to  Multiplication  and  Division,   ....  39 

Compound  Numbers, 48 

Reduction, 54 

Compound  Addition, 60 

Application  of  Compound  Addition, 64 

Compound  Subtraction, 65 

Compound  Multiplication, 68 

Compound  Division, 71 

SECTION  III. 

Vulgar  Fractions, 76 

Complex  Fractions, 83 

Reduction  of  Fractions, 85 

Addition  of  Fractions, 92 

Subtraction  of  Fractions, 94 

Multiplication  of  Fractions, 96 

Division  of  Fractions, 97 

Decimal  Fractions, 100 

Addition  of  Decimals, • 102 

Subtraction  of  Decimals, 103 

Multiplication  of  Decimals, 104 

Division  of  Decimals, , 106 

Reduction  of  Decimals,    ...  * 110 

SECTION  lY. 

Comparison  of  Numbers,   ...         116 

Proportion,  or  Rule  of  Three,  119 


X  COIS'TE^Tg. 


Application  of  Proportion, • 122 

Compound  Proportion, 134 

Practice, .  • 143 

Simple  Interest, 151 

Interest  on  Notes,  Bonds,  Slc, 159 

Compound  Interest, 166 

Discount  and  Banking, 171 

Per-Centage, 176 

Stocks, 179 

Equation  of  Payments, 181 

Profit  and  Loss  per  cent., 184 

Reduction  of  Currencies, 188 

Exchange, 192 

Exchange  and  Per-Centage  combined, 195 

Conjoined  Proportion, ' 199 

Fellowship, 203 

Taxes, 205 

Compound  Fellowship, • 207 

Barter, 210 

SECTION  V. 

Mensuration, 212 

Mensuration  of  Solids, 219 

Involution  of  Powers, 226 

Square  Root, 230 

Application  of  Square  Root, 235 

Problems  on  the  right-angled  Triangle, 238 

Cube  Root, 241 

Application  of  Cube  Root, 247 

Supplement  to  Square  and  Cube  Roots, 249 

Abbreviations  in  Cube  Root, 251 

Approximate  Cube  Roots  of  Surds,    ........  254 

Roots  of  all  Pov/ers, 258 

Arithmetical  Progression, 259 

Geometrical  Progression, 262 

Alligation, 266 

Position, 269 

Permutation, 272 

Combination, 273 

Miscellaneous  Examples, 274 

Appendix, 280 


ARITHMETIC 


Arithmetic  is  the  science  of  numbers.  It  would  be 
evidently  impossible  to  express  each  number  by  a  sepa- 
rate character ;  for  in  that  case  we  must  have  an  infinite 
number  of  characters. 

For  example,  to  express  every  number,  from  one  up 
to  one  million,  we  must  have  a  million  of  different  signs 
or  marks,  unless  we  can  accomplish  the  end  by  changes 
and  combinations  of  a  few  simple  characters. 


(Art.  1.)  Numbers  were  at  first  written  separately, 
either  in  words  at  length,  or,  as  among  the  Romans,  in 
characters,  commonly  the  letters  of  the  alphabet ;  and  for 
larcje  numbers,  such  letters  were  used  as  would  make  up, 
by  the  addition  of  their  separate  values,  the  number  re- 
quired. Thus,  the  number  four  hundred  and  fifly-eight 
was  written  CCCCLVIIL,  the  value  of  C  being  ©ne  hun- 
dred, L  fifty,  V  five,  and  I  one. 

The  system  of  notation  now  in  use,  and  which  origin- 
ated in  Arabia,  is  so  contrived  as  to  express  all  numbers 
with  ten  characters  only.  Nine  of  these  are  significant, 
and  represent  numbers,  and  the  other  is  used  to  denote 
nothing,  or  the  absence  of  quantity. 

To  express  numbers  greater  than  nine,  recourse  is  had 
to  a  law  which  assigns  different  values  to  the  figures,  ac- 
cording to  the  position  which  they  occupy.  According 
to  this  law,  uniis  of  the  first  order  occupy  the  first  place 
on  the  right  of  the  written  expression,  units  of  the  second 
order,  the  second  place,  and  so  of  the  others. 

The  numbers  1,  2,  3,  4,  5,  6,  7,  8,  9,  are  each,  as  we 
perceive,  separate  characters ;  but  when  we  would  write 

n 


12  ARITHMETIC. 


ten,  which  is  a  unit  of  the  second  order,  M^e  place  unity  in 
the  second  place,  and  a  0  (cypher)  in  the  first,  thus,  10, 
which  may  either  be  considered  as  one  unit  of  tlie  second 
order,  or  ten  units  of  the  first.  Proceeding  with  units  of 
the  second  orner  in  the  same  manner  as  with  those  of  the 
first,  we  shall  have  10,  20,  30,  40,  50,  60,  70,  80,  90,  and 
next  a  unit  of  the  third  order,  which  is  written  100,  the 
significant  ligure  being  written  in  the  third  place.  In  like 
manner,  we  pass  from  the  third  to  the  fourth  order. 

(Art.  2.)  To  form  an  adequate  and  correct  conception 
of  large  numbers,  M-e  must  have  names  for  the  difierent 
orders  of  places  from  the  right  hand,  which  will  enable  us 
to  read  off"  the  numbers  in  words.  This  naming  and 
reading  is  called 

NUMERATION. 

The  first  order  of  numbers  is  called Units. 

The  second  order, Tens. 

The  third  order, Hundreds. 

The  fourth  order, Thousands. 

The  fifth  order, Tens  of  Thousands. 

The  sixth  order, Hundreds  of  Thousands. 

The  seventh  order, Millions. 

The  eighth  order, Tens  of  Millions. 

The  ninth  order, Hundreds  of  Millions. 

The  tenth  order, Billions. 

The  eleventh  order, Tens  of  Billions. 

The  twelfth  order, Hundreds  of  Billions. 

The  thirteenth  order, Trillions. 

And  we  may  thus  go  on,  and  repeat  the  tens  and  hun- 
dreds of  trillions,  and  then  take  another  name  ;  and  again 
and  again  repeat  the  tens  and  hundreds  belonging  to  the 
ne^v  name. 

Hence  v/e  observe,  that  we  can  divide  these  orders  in- 
to periods  of  three  figures  each,  and  give  to  each  period 
its  distinctive  name,  reading  ofl^  the  hundreds,  tens  cmd 
uni's,  in  each  period. 


NUMERATION.  13 


w        c         c 

§    -^     -I 


KXAMPLE. 


3         S       "S       r=: 


80.652.941.600.807.362.546.278.009.650.208. 

Which  is  read,  eighty  nonillions,  six  hundred  fifty-two 
octiUions,  and  so  on  to  the  last  period,  to  which  the  name 
(units)  is  not  added. 

Divide  off  by  points,  and  read  the  following  numbers. 

864321 

76247523 

9210043747829 

100210031002478298 

To  write  numbers  from  words,  we  begin  on  the  left 
hand,  taking  care  to  fill  up  those  periods  or  places  that 
are  omitted  in  the  question. 

Write  the  following  numbers  : 

Nine  millions,  seventy-two  thousand,  and  two  hun- 
dred. 

Eight  hundred  millions,  forty-four  thousand,  and  fifty- 
five. 

Eight  billions,  sixty-five  millions,  three  hundred  and 
four  thousand,  and  seven. 

Fifty-four  sextillions,  three  hundred  trillions,  sixty-se- 
ven millions,  four  hundred  and  twenty. 

Seventy  decillions,  two  hundred  and  thirty-one  octil- 
lions, one  billion,  one  hundred  thousand,  and  three  hun- 
dred. 

Six  hundred  and  forty  thousand,  four  hundred  and 
eighty-one. 

Three  millions,  two  hundred  sixty  thousand,  one  hun- 
dred and  six. 

From  the  system  of  notation  already  explained,  it  is 
evident  that  figures  have  a  simple  value  and  a  local  value. 

When  a  figure  stands  in  the  place  of  units,  it  has  a 
simple  value  only. 

B 


14  ARITHMETIC. 


THE  FOUR  RULES  OF  ARITHMETIC. 

(Art.  3.)  As  quantity  admits  of  no  other  changes  than 
increase  and  diminution,  it  is  evident  that  all  the  operations 
of  arithmetic  must  be  based  upon  these  only. 

Under  tlie  first  are  comprehended  addition  and  multi- 
plication, and  under  the  second,  subtraction  and  division. 

Addition  consists  in  finding  the  sum  of  two  or  more 
numbers. 

Multiplication  is  the  successive  addition  of  a  number 
to  itself  a  given  number  of  times. 

Subtraction  consists  in  finding  the  difference  between 
two  numbers,  or  diminishing  one  given  number  by  an- 
other. 

Division  is  the  same  as  subtracting  one  number  succes- 
sively from  another,  in  order  to  find  how  many  times  the 
smaller  number  is  contained  in  the  larger. 

Addition,  subtraction,  multiplication  and  division,  are 
called  i\\efour  rules  of  arithmetic  ;  all  arithmetical  oper- 
ations whatever  are  combinations  of  these. 

(Art.  4.)  In  arithmetical,  and  in  all  subsequent  math- 
ematical operations,  characters  and  signs  are  used  to  de- 
note operations  and  conditions. 

The  perpendicular  cross,  thus  +»  denotes  addition. 

The  horizontal  line,  thus  — ,  denotes  subtraction. 

The  diamond  cross,  or  a  point,  thus,  X  »  (either  one), 
indicates  multiplication. 

A  horizontal  line,  with  a  point  above  and  below,  thus 
-^,  indicates  division.  Points,  thus,  :  :;  or,  :=:  pro- 
portion.    Double  horizontal  lines,  thus  =,  equality. 

The  following  character  represents  square  root,  J. 

The  same  character,  M'ith  a  small  figure  annexed,  as  3, 
4,  5,  &;c.,  thus  ^J,  "^J,  ^J,  indicates  third,  fourth  and 
fifth  roots. 

(Art.  5.)  Numbers  are  referable  to  things  ;  but  when 
no  reference  is  made,  the  number  is  said  to  be  abstract ; 
for  instance,  tlie  number  3  is  abstract,  and  can  be  applied 


THE    FOUR    RULES    OF    ARITHMETIC.  15 


to  any  object  whatever,  as  three  apples,  three  horses, 
three  dollars,  &c.;  and  when  it  is  so  applied,  it  is  no  long- 
er abstract. 

Abstra(;t  numbers  can  be  added  together,  as  they  are 
then  understood  to  refer  to  the  same  thing  or  scale  of 
measure,  but  numbers  referring  to  different  things,  or 
different  scales  or  standards  of  measure,  cannot  be  put 
into  one  sum. 

Let  the  pupil  early  imbibe  this  important  idea,  that 
numbers  referring  to  different  things,  cannot  be  added  to- 
gether, under  any  rule,  nor  in  any  part  of  arithmetic. 
For  example  :  3  apples  and  5  apples  are  8  apples  ;  but 
3  apples  and  5  dollars  are  not  8  of  the  one  or  the  other, 
and  it  is  manifest  that  they  cannot  be  added  together. 

As  an  objection  to  this,  a  person  might  say,  that  in  a 
certain  orchard  there  are  23  apple-trees,  5  pear-trees,  and 
7  cherry-trees  ;  how  many  in  all  ? 

We  certainly  cannot  add  them  together  as  apple-trees, 
because  ihey  are  not  all  apple-trees,  nor  are  they  all  pear- 
trees  or  cherry-trees  ;  but  we  can  add  them  together  un- 
der the  general  name  of  trees. 

In  the  same  way,  in  estimating  a  farmer's  stock,  we 
may  add  together  horses,  oxen,  cows,  sheep,  &c.,  under 
the  general  name  of  animals. 

Indeed  we  may  go  further,  and  say,  In  a  certain  field 
I  saw  6  trees,  3  hay-stacks,  and  2  oxen  ;  how  many  ob- 
jects were  before  my  vision  1 

Here  we  may  add  together  trees,  hay-stacks,  and  oxen, 
or  throw  them  into  one  sum,  under  the  general  name  of 
objects. 

Yet  the  first  assertion  holds  good,  that  we  cannot  add 
numbers  together  unless  they  refer  to  one  common  stand- 
ard. In  the  first  observation  above  cited,  we  may  put  the 
numbers  together  as  trees,  in  the  second,  as  animals,  in 
the  third,  as  objects. 

With  these  preliminary  remarks,  we  commence  with 
tbe  first  arithmetical  operation. 


16  ARITHMETIC. 


ADDITION. 

(Art.  6.)  Addition  is  the  art  of  collecting  several  quantities  into 
one  sum.  We  illustrate  this  definition  by  the  few  following  men- 
tal exercises. 

1.  A  man  bought  a  load  of  wood  for  4  dollars,  a  barrel  of  flour  for 
5  dollars,  and  a  barrel  of  pork  for  13  dollars;  how  many  dollars  did 
he  spend  ? 

2.  A  man  gave  9  apples  to  one  boy,  to  another  8,  and  to  another 
10  ;  how  many  apples  did  he  give  away  1 

3.  One  boy  has  11  cents,  another  has  7  cents,  and  another  has  10 
cents  ;  how  many  cents  have  the  3  boys  ? 

In  this  last  example,  shall  11,  7,  10  and  3  be  added  together;  and 
if  not,  why  ] 

But  all  numbers  cannot  be  added  mentally,  because  of  their  mag- 
nitude ;  therefore  we  must  have  a  more  artificial  rule ;  and  as  unlike 
things  cannot  be  added  together,  preparatory  to  addition  we  must  write 
down  the  dilferent  sums,  so  that  units  shall  stand  under  units,  tens 
under  tens,  hundreds  under  hundreds,  <S^c.,  ^-c,  then  the  student  can 
recognise  the  rationale  of  the  following 

Rule.  Commence  with  the  right  hand  column,  mentally  Jin  d  its 
sum,  and  ivrite  the  unit  figure  of  that  sum  directly  under  the  col- 
umn ;  add  up  the  next  column  in  the  same  manner,  and  increase  it 
by  the  tens  from  the  last  column,  the  unit  of  this  sum  write  under 
the  second  column,  and  carry  its  tens  to  the  next  column ,-  and  so 
continue,  from  column  to  column,  to  the  last  one,  under  which  its 
whole  sum  mu^t  be  put  doicn. 

The  reason  for  carrying  one  for  every  ten,  from  column  to  column, 
is  because  we  have  ten  characters,  and  hence,  if  a  sum  is  more  than 
ten,  it  cannot  be  expressed  by  any  single  character,  and  a  ten  of  any 
order  of  numbers  makes  orie  of  the  next  higher  order. 

We  illustrate  this  by  the  following  example,  done  out  in  full,  set- 
ting down  the  sum  of  each  column  in  its  proper  place  : 

Here  the  reason  of  adding 
the  one  of  the  14  to  the  next 
column,  is  clearly  seen,  by  in- 
spection. 


Thus 

4372 

5487 

3445 

Sum  of  the  units, 

14 

"    "     "    tens, 

19 

"    "     "    hundreds. 

11 

"    "     "    thousands, 

12 

Whole  sum,  13304 

After  the  principle  is  understood,  we  do  not  spread  the  work  out,  as 
above,  but  mentally  add  the  tens  of  one  column  into  the  next. 

If  this  mental  process  is  troublesome,  and  there  is  danger  of  losing 
the  number  to  be  carried,  the  pupil  can  write  units  under  the  first 


ADDITION.  17 


column,  and  the  tens  over  the  next  column,  which  will  become  a  part 
of  it :  and  so  on. 
Tims,      121  2  2  111 

4373  (478 

.5487  Add  <  389  Add 

3445  (473 


Sum,  13304  1340  17776 


(I)  (2)  (3)  (4) 
432  9376  27636  84736 
213  810  79892  78928 
121  57  38941  146469  27890 
213  973  67832  33217 
92  59244  127076      47263 


Sum  979 


Sum  11308     273545     273545 

N.  15. — There  is  no  simple  proof  of  addition  *  unless  we  call  a  different 
orfler  or  manner  of  putting  the  sums  together  proof,  in  case  the  final  sums 


*  We  say,  no  simple  proof  of  addition.  There  is  a  proof  by  casting  out 
the  9's,  but  its  principle  is  too  refined  and  scientific  to  be  comprehended  by 
persons  just  learning  the  science  of  numbers;  we  therefore  give  it  in  a  note. 
The  rule  is  as  follows: 

Add  together  the  figures  in  the  upper  horizontal  line,  reject-    Exa7nple. 
ing  the  9's  and  writing  down  the  excess  of  9  a  little  to  the  Excesses. 

right.  Do  the  sam^  with  every  horizontal  column.  Add  up  3497  5 
these  excesses  of  9,  thus  found,  and  reject  the^9  from  their  6512  5 
sum;  and  if  the  work  is  right,  this  last  excess  of  9's  is  the       8-.295    6 

same  as  the  excess  of  9's  from  the  sum  total.     Thus,  in  our       

example  the  sum  of  5,  5,  6.  is  16 ;  excess  over  9  is  7,  and  the     18304    7 
excess  of  9's  in  18304  is  also  7.    There  is  a  possibility  of  the  work  proving 
right,  and  still  being  wrong;  but  it  is  barely  a  possibility.    Thus,  if  we  had 
made  the  sum  total  18403,  it  would  be  wrong,  but  the  excess  over  9's,  taken 
horizontally,  is  still  7. 

This  rule  is  founded  on  the  following  property  of  numbers,  namely  :  .Any 
number  divided  by  9  icill  leave  the  same  remainder  as  the  sum  of  its  digits  di- 
vided by  9.  Thus,  2S  divided  by  9  leaves  a  remainder  of  1,  and  2  added  to 
8  gives  10,  which,  divided  by  9,  also  leaves  1 :  and  so  with  any  other  num- 
ber. But,  to  prove  this  generally,  we  may  take  any  number,  say  3476,  and 
decompose  it  thus: 

/  3000=3(1000)=3(999)-f3 

)     400=4(   100)=4(  99)-f  4 

^^^^  \       70=7(     10)=7(     9)-l-7 

(         6=  6  =  6 

Here,  then,  we  perceive  by  inspection  that  3000  has  a  certain  number  of 
9's,  and  a  remainder  of  3 ;  400  has  a  certain  number  of  9's,  and  a  remainder 
of  4;  70  has  a  certain  number  of  9's,  and  a  remainder  of  7;  and  6,  being  un- 
der 9,  is  called  a  remainder.    But  these  remainders,  written  horizontal- 
— 


18 


ARITHMETIC. 


a^ree;  but  tliis  gives  no  demonstraiion  lliat  an  «-rror  h;is  not  hcen  made. 
Thus,  in  the  (3)  example  we  added  up  ihe  eiiiire  cohunns  in  the  usual  man- 
ner. Tlieu  we  took  the  sum  of  several  parts,  and  the  .sum  of  these  parts,  as 
may  be  seen  by  inspeclioii.  and  tiie  final  snm.s  aijree  ;  hut  if  an  error  was 
made  in  the  first  instance  in  adding  the  two  upper  or  two  lower  sums, 
tiie  same  error  might  be  made  in  the  secoiul  instance. 


(•^) 

(6) 

(7) 

7298104.3 

1278976 

416785413 

67126459 

7654301 

915123460 

39412767 

876120 

31810213 

7891234 

723456 

7367985 

109126 

31309 

654321 

84172 

4871 

37853 

72120 

978 

2685 

187676921 

10570011 

1371781930 

8.  Add  8635,  2190,  7421,  5063,  2196,  and  1245  together. 

A71S.  26750. 

9.  Add  246034,  298765,  47321,  58653,  64210,  5376,  9821,  and 
340  to<?ether.  .4/?.?.  730520. 

10.  Add  27104,32540,  10758,6256,704321,730491.  2787316, 
and  2749104  together.  Ans.  7047890. 

11.  Add  1,  37,  29504,  6790312,  18757421,  and  265  together. 

Ans.  25577540. 

12.  Add  562163,  21964,  56321,  18536,  4340,  279,  and  73  toge- 
ther. Am.  663676. 

13.  What  is  the  sum  of  the  following^  numbers?  viz:  Seventy- 
five  ;  one  thousand  and  ninety-five  ;  six  thousand  four  hundred  and 
thirty-five;  two  hundred  and  sixty-seven  thousand;  one  thousand 
four  hundred  and  fifty-fiye  ;  twenty-seven  millions  and  eighteen  ;  two 
hundred  and  seventy  millions  and  twenty-seven  thousand. 

Ans.  297303078. 

APPLICATION. 

1 .  Add  3742  bushels,  493  bushels,  927  bushels,  643  bushels,  and 
953  bushels  together.  Ans.  6758  bushels. 

2.  Add  7346  acres, '  9387  acres,  8756  acres,  8394  acres,  and 
32724  acres.  Ans.  66607. 

3.  Henry  received  at  <me  time  15  apples,  at  another  115,  at  ano- 
ther 19;  how  many  did  he  receive]  Ans.  149. 

4.  A  person  raised  in  one  year  724  bushels  of  corn,  in  another 
3498  bushels,  in  another  9872;  how  much  in  all] 

A)is.  14094  bushels. 


ly  3476,  give  the  original  number ;  therefore,  Uie  ssnm  of  the  digits  of  any 
number,  divided  by  9,  will  give  the  same  reir.'.under  as  the  lumiber  divided 
by  9,  which  was  to  be  demonstrated. 

Now,  tliis  property  bemg  demonstrated,  the  reason  of  the  rule  is  obvious ; 
for  the  excess  of  9's  in  the  several  sums  added  together,  and  their  excess 
over  9'3  being  taken,  must  be  equal  to  the  excess  of  9's  in  the  sum  total, 
as  the  sum  of  all  the  parts  of  a  number  must  equal  the  whole. 


ADDITION. 


5.  A  man  on  a  journey  traveled  the  first  day  37  miles,  the  second 
33  miles,  the  third  40  miles,  the  fourth  35  miles  ;  how  far  did  ho  travel 
in  the  four  days  ]  Atjs.  145  miies. 

6.  A  has  a  flock  of  sheep  containing  34,  B  has  a  flock  of  47,  and 
U  a  flock  of  54 ;  how  many  sheep  are  there  in  the  three  floclcs  ? 

A77S.  135  sliecp. 

7.  The  distance  from  Philadelphia  to  Bristol  is  20  miles,  from 
Bristol  to  Trenton  10  miles,  from  Trenton  to  Princeton  12  miles, 
from  Princeton  to  Brunswick  18  miles,  from  Brunswick  to  IVevv 
York  30  miles  ;  how  many  miles  from  Philadelphia  to  New  York  ? 

A)}s.  90  miles. 

8.  A  person  bought  of  one  merchant  10  barrels  of  flour,  of  ano- 
ther 20  barrels,  of  another  95  barrels;  how  many  barrels  did  he  buy  ? 

Ans.  125  barrels. 

9.  A  wine  merchant  has  in  one  cask  75  gallons,  in  another  65,  in 
a  third  57,  in  a  fourth  83,  in  a  fifth  74,  and  in  a  sixth  67  gallons  ; 
how  many  gallons  has  he  in  all?  A?is.  421. 

10.  An  estate  is  to  be  shared  equally  by  four  heirs,  and  the  portion 
to  each  heir  is  to  be  3754  dollars;  what  is  the  amount  of  the  estate? 

A>is.  15016  dollars. 

11.  A  certain  regiment  of  soldiers  consists  of  eight  companies,  and 
each  company  consists  of  80  men ;  how  many  soldiers  are  there  in 
the  regiment?  Ans.  640. 

12.  How  many  men  are  there  in  an  army  consisting  of  52714  in- 
fantry, 5110  cavalry,  6250  dragoons,  3927  light  horse,  928  artillery, 
250  sappers,  and  406  miners  ?  Ans.  69585. 

13.  A  merchant  deposited  56  dollars  in  a  bank  on  Monday,  74  on 
Tuesday,  120  on  Wednesday,  96  on  Thursday,  170  on  Friday,  and 
50  on  Saturday  ;  how  much  did  he  deposit  during  the  week  ] 

Arts.  466  dollars. 

1 4.  A  merchant  bought  at  public  sale  746  yards  of  broadcloth,  050 
yards  of  muslin,  2100  yards  of  flannel,  and  250  yards  of  silk ;  how 
many  yards  in  all  ?  Ans.  3746  yards. 


FEDERAL  MONEY. 

(Art.  8.)  This  is  the  name  given  to  the  established  currency  of 
the  United  States  ;  and  it  is  purposely  adjusted  to  run  on  the  same 
scale  of  computation  as  abstract  numbers :  that  is,  fen  of  one  order 
making  one  of  the  next  superior  order.  Therefore,  this  money 
must  be  added  by  the  same  rule  as  whole  or  abstract  numbers.  Its 
denominations  are  Eagles,  Dollars,  Dimes,  CeJits,  and  Mills.* 
Eagles  are  the  highest,  and  mills  the  lowest.  Ten  mills  make  a  cent, 
ten  cents  make  a  dime,  ten  dimes  make  a  dollar,  and  ten  dollars 
make  an  eagle.     In  practice,  mills,  dimes,  and  eagles  are  disregarded 

*The  coins  of  the  United  States  are  eagles,  half-eagles,  and  quarter-eagles, 
of  gold  ;  the  doUnr,  hall'-dollar.  quarter-dollar,  dime,  an  half-dime,  of  silver ; 
the  cent  and  halt'-cent  of  copper. 


20  ARITHMETIC. 


in  naming  money,  and  we  have  dollars  and  cents  only.  The  eagles 
are  read  ^s  dollars,  the  dimes  as  cents;  and  the  mills  are  quite  ne- 
glected in  aggregate  sums.  The  dollar  is  considered  as  unity  ;  its 
mark  is  $,  and  a  point.,  called  i\\e  decimal  point,  is  placed  at  the  right 
hand  of  the  dollars  to  separate  them  from  the  inferior  denominations. 
'J''hus,  twelve  dollars  and  seventy-five  cents  is  written,  $12.75.  In 
reading,  if  all  the  denomination?  were  named,  we  should  say  one  eagle, 
two  dollars,  seven  dimes,  and  five  cents;  but  this  would  he  too  formal 
and  ijiconvenient.  We  may  change  the  unit  by  removing  the  deci- 
mal point;  thus,  in  the  above  sum,  twelve  dollars  and  seventy-five 
cents  may  be  considered  1275  cents,  the  point  only  is  removed,  and  a 
cent  is  now  unity  in  place  of  a  dollar. 

The  following  rule  for  addition  in  federal  money  corresponds  to  the 
dollar  being  unity  :  d  stands  over  dimes,  c  over  cents,  and  m  over 
mills;  but  in  practice  these  are  not  written,  for  the  localily  from  the 
decimal  point  determines  the  value  and  denomination  of  the  figure. 

RuLK.  rhtce  dollars  under  dollars,  cents  under  cents,  SfC,  and 
add  as  in  whole  numbers  ,-  and  put  the  separatrix  in  the  sum  to- 
taL  under  the  separating  points  above. 

If  the  cents  are  less  than  10,  a  cypher  must  be  put  in  the  place 
of  dimes. 


(0 

(2) 

(3) 

(4) 

$  d  c  m 

$  d  c 

d  0  m 

$  d  c  m 

139.8  6  7 

4319.8  9 

.9  7  6 

81.0  5  3 

1273.5  9  4 

3287.8  0 

.4  5  8 

67.4  I  2 

9807.(i  0 

1829.1  6 

.6  2  9 

37.4  5  3 

7695.2  4 

130.0  1 

.5  8  3 

21.6  5  3 

$18916.3  0  1 

207.5  7  1 

5.  What  is  the  sum  of  140  dollars  9  cents,  24  dollars  16  cents,  5 
dollars  25  cents,  5  dollars  5  cents,  and  304  dollars  8  cents  and  9  mills. 

Ans.  $478,639. 

6.  A  man  left  his  estate  to  his  two  children,  who,  after  payment  of 
the  debts,  amounting  to  645  dollars  75  cents,  received  each  3842  dol- 
lars ;  how  much  was  the  whole  estate  ?  Ans.  $8329.75. 

7.  If  I  owe  to  A  50  dollars  and  8  cents,  to  B  12  dollars  more  than 
that  sum,  and  to  C  75  cents  more  than  to  both  the  other.s,  how  much 
do  I  owe  them  all  ?  Ans.  $265.07. 

8.  A  farmer  purchased  a  cow  for  $14.75,  a  horse  for  $75.14,  a  wag- 
on for  $31,375,  and  a  harness  for  $41,625;  what  was  the  whole 
amount?  ^/7.s.  $161.89. 

9.  My  grocer  sends  me  the  following  bill :  To  2  pounds  of  coflfee  at 
12  cents  5  mills  per  pound;  a  roll  of  butter  for  87|  cents;  and 
a  box  of  soap  for  2  dollars  and  65  cents;  what  is  the  amount  of 
the  bill  ?  Ans.  $3,775. 

10.  What  is  the  sum  of  25  eagles,  62  dollars,  8  dimes,  75  cents, 
and  5  mills,  properly  expressed  in  dollars  and  cents  ? 

Ans.  $313.55^. 


ADDITION.  •       21 


Quesf/oris. 

How  many  primary  rules  of  arithmetic  are  there? 

W'liat  are  they  called  ? 

What  is  addition  ? 

How  do  you  place  nunihers  to  be  added? 

Whore  do  you  begin  the  addition? 

Why  do  you  carry  one  for  ten,  in  preference  to  any  other  number  ? 

W'hy,  at  the  last  column,  do  you  write  the  whole  number? 

Why  is  federal  money  added  as  natural  or  abstract  numbers  ? 


SUBTRACTION. 

(AnT.  9.)  "SiMPLK  SuKTnACTiON  is  taking  a  less  number  from 
a  greater,  and  noting  the  difference  or  remainder;  thus,  6  taken  from 
10,  theditierence  is  4;  and  5  dollars  taken  from  7  dollars,  leaves  a  re- 
mainder of  2  dollars.  By  signs,  these  examples  may  be  expressed 
thus:    10—6=4;  7—5=2. 

"  The  larger  of  the  two  numbers  is  called  the  minuend,  the  lesser 
is  called  the  subtrahend,  and  the  difference  is  called  their  remainder.''^ 

JIEiXTAL  EXEnciSES. 

1.  Charles  had  25  cents,  and  paid  10  of  them  for  a  book  ;  how 
many  had  he  left  ? 

2.  A  merchant,  35  years  old,  had  been  in  trade  13  years ;  at  what 
age  did  he  commence  business  1 

3.  A  father  who  was  37  years  of  age,  had  a  son  25  years  younger; 
how  old  was  the  boy  ? 

4.  A  horse  and  harness  cost  183  dollars.  The  harness  cost  84 
dollars  ;  what  was  the  price  of  the  horse  1 

5.  Dr.  Franklin  died  in  1790,  at  the  age  of  84  ;  in  what  year  was 
he  born  ? 

6.  A  man  paid  84  dollars  for  a  watch,  but  was  obliged  to  sell  it  for 
07  dollars  ;  how  many  dollars  did  he  lose] 

NoTK. — All  operations  in  both  addition  and  sublraction  g^o  back  to  the 
mental  proce.ts ;  hut  when  results  are  written  down  as  last  as  they  are  at- 
tained, it  is  called  written  arithmetic.  And  the  art  of  writing  comes  in  to 
help  ihe  mental  process,  and  enables  the  mind  to  concentrate  its  powers  on 
a  sing^le  point  at  atime. 

liCt  ns  write  the  last  problem,  and  submit  it  to  a  more  artificial  process; 
but  when  written,  if  we  consider  only  one  figure  at  a  time,  we  cannot  sub- 
tract 7  from  4.     We  must,  therefore,  decompose  (he  numbers,  thus  : 

84=70-1-14 
67=60-f-  7 


Difference,  I0-|-  7,  or  17. 

Again,  let  us  subtract  765  from  1234.    To  analyze  this,  1234  must  be  so 


22 


ARITHMETIC. 


separated  tliat  it  would  be  possible  to  subtract  7  liundreds,  6  tens,  and  5 
uuils  from  it,  thus: 

1234=1100-1-1204-14 

765=  700-}-  60-f-  5 

Dlff  rence,    4tj9=  400-|-  GO-j-  9 

"\Vc  decompose  the  numbers  to  enable  the  pupil  to  perceive  the  raiionale 
of  llie  vviiole  operation.  Thus,  it  will  be  observed  that  we  must  take  5  uiiiis 
from  14  uiii:s;  but,  to  make  the  4,  14.  we  must,  (as  it  is  called), borrow  a  ten, 
or  one  iVom  the  tens.  Then  it  will  be  G  I'rom  2.  or  12,  (but  in  place  of  6  from 
12.  we  mav  say  7  from  13.  the  difference  is  the  same).  Tlien,  7  fioiri  11,  or 
wiiicli  gives  the  same  difference,  8  I>om  12.  4.  Hence,  when  we  subtract  a 
figure  irom  a  less  one  above,  we  conceive  tlie  upper  figure  increased  by  10, 
and  the  next  figure  in  the  lower  line  increased  by  one  to  compensate  it. 

From  these  observations  we  draw  the  following 

liuLK,  To  operate  on  large  numbers,  set  doivn  the  less  niunher 
under  the  greater,  units  under  units,  tens  under  tens,  S(C.,  and  draw 
a  line  htntath  them. 

Then,  beginning  at  the  right  hand,  subtract  each  figure  from  the 
one  directly  over  it,  and  set  down  the  remainder. 

Bat  if  the  upper  figure  be  the  least,  suppose  it  to  be  increased  bij 
ten  ;  then  make  the  subtraction,  set  down  the  rernainder,  and  carry 
one  to  the  next  figure  of  the  subtrahend. 

Pkoof.  Add  the  remainder  to  the  subtrahend.  If  their  sum  is 
equal  to  the  minuend,  the  work  may  be  regarded  as  right. 


EXAMPLES. 


Minuends, 
Subtrahends, 

Remainders, 

Proofs, 


859267 
107895 

751372 

859267 


(4) 
10000 
4 


(5) 

3000 

999 


(2) 

6794213 

975678 

5818535 

6794213 

(6) 
6798 
4000 


(3) 

19067803 

04202196 


148G5607 


19067803 


(7)      (8) 
10000    87000 
1     1009 


Remainders,        9996         2001         2798 


9999 


85991 


9.  Ft 

10.  From 

11.  — 

12.  — 

13.  — 

14.  — 

15.  — 

16.  — 

17.  — 


78213609  take  27821890. 


Ans.  50391719. 


196 

487 

875 

967 

1001 

9765 

87696 

455692 


take 


37. 

96. 
302. 
351. 

487. 

1307. 

10091. 

300120. 


Resul 


159 
391 
573 

616 

514 

8'158 

77605 

155572 


SUBTRACTION.  23 


18.  From   eleven   thousand   and  eleven  Ijumlred  take.  1521). 

liciii.  10580. 

19.  From  thirty  thousand  and  ninely-scvi-n,  take  one  thousand  six 
hundred  and  lilty-four.  Ktm.  2841:3. 

20.  From  one  hundred  million  two  hundred  and  lorty-.seven  thou- 
sand, take  one  million  four  hundred  and  nine.       Kern.  9924G591. 

21.  Subtract  one  from  one  million.  Rem.  999999. 

APPLICATION. 

1.  Sir  Isaac  Newton  was  born  in  the  year  1642,  and  died  in  1727  ; 
to  what  age  did  he  live?  Ana.  85  years. 

2.  A  merchant  gave  his  note  for  5200  dollars.  He  paid  at  one 
time  2500  dollars,  at  another  time  175  dollars  ;  what  remained  due  ? 

Ans:  2525  dollars, 

3.  -In  1800  there  were  903  post-offices  in  the  United  States,  and  in 
1828  there  were  7530 ;  how  much  had  their  number  increased  ? 

Aus.  6627. 

4.  A  farmer  owned  360  acres  of  land,  and  sold  175  acres;  how- 
many  acres  had  he  left!  ^1"^.  185. 

5.  Massachusetts  contains  7800  square  miles,  and  New  Hampshire 
9491  square  miles;  which  is  the  largest,  and  by  how  many  square 
miles  ? 

6.  New  York  contains  46085  square  miles  ;  how  much  larger  is  it 
than  Massachusetts  ?  than  New  Hampshire  ?  than  both  these  States 
together ] 

7.  The  Declaration  of  Independence  was  published  July  4th,  1776  ; 
how  many  years  to  July  4th,  1850? 

8.  Gunpowder  was  invented  in  the  year  1330  after  Christ,  and  the 
art  of  printing  in  the  year  1441  ;  how  long  from  each  of  those  events 
to  the  year  1850? 

9.  The  mariner's  compass  was  invented  in  Europe  in  the  year 
1302  ;  how  long  since  that  event  to  the  present  year  ? 

10.  A  man  deposited  in  bank  8752  dollars,  and  drew  out  at  one 
time  4234  dollars,  at  another  1700  dollars,  at  another  962  dollars, 
and  at  another  49  dollars ;  how  much  had  he  remaining  in  bank  1 

Ans.  1807  dollars. 

11.  A  merchant  bought  5875  bushels  of  wheat,  and  sold  2976 
bushels',  how  many  bushels  remain  in  his  possession  ? 

Ans.  1899  bushels. 

12.  A  grocer  bought  25  hogsheads  of  sugar,  containing  250  hun- 
dred weight,  and  sold  9  hogsheads,  containing  75  hundred  weight; 
how  many  hogsheads  and  how  many  hundred  weight  had  he  left. 

Ans.  16  hogsheads  and  175  hundred  weight. 

13.  A  traveler  who  was  1300  miles  from  home,  traveled  homeward 
235  miles  in  one  week,  in  the  next  275  miles,  in  the  next  325  miles, 
and  in  the  next  280  miles;  how  far  had  he  still  to  go  before  he  would 
reach  home?  Ans.  175  miles. 


24  ARITHMETIC. 


SUBTRACTION  OF  FEDERAL  MONEY. 

(Art.  10.)  Rule.  Place  dollars  under  dollars,  cents  under 
cents,  Ac- ;  proceed  as  in  whole  numbers,  and  put  the  separatrix  in 
the  remainder,  under  the  decimal  points  above. 

(I)  (2)  (3) 

$  c  $cm  $cm 

From      75.48  246  05  6  6327.86  5 

Take       58.76  218.60  7  4961. 


Ans.      §16.72 


(2) 

$    c 

m 

246  05 

6 

218.60 

7 

4.  From  36  dollars  and  5  cents,  take  28  dollars  and  60  cents. 

Rem.  $7A5. 

5.  From  two  dollars,  take  5  dimes  and  5  mills.       Rem.  §1.49^. 

6.  From  2  dollars,  take  7  cents.  Rem.  $1.93. 

7.  From  4  dollars,  take  75  cents.  Rerii.  $3.25. 

8.  Owing  my  neighbor  60  dollars,  I  paid  him  at  one  time  §9.50  ; 
at  another  5il5;  at  another  $30.75;  and  at  another  62^  cents ;  how 
much  was  then  due  him?  Ans.  §4.12^. 

9.  A  man  paid  for  two  sheep  7  dollars  and  50  cents,  and  for  a  cow 
twice  as  much,  lacking  75  cents;  how  much  did  she  cost  him  ? 

An>:  $14.25. 

10.  A  farmer's  bill  at  a  store  amounted  to  $37,625,  and  he  paid 
$19  and  5  cents  in  oats ;  how  much  did  he  still  owe  ? 

Ans.  $18,575. 

1 1.  A  man  owes  §100  :  he  has  §47.375 ;  how  much  must  he  bor- 
row to  pay  the  debt?  Ans.  §52.625. 

12.  A  man  sold  his  house  and  lot  for  §2070,  which  was  §62.875 
more  than  it  cost  him ;  what  did  it  cost  him  ]         Ans.  §2007.125. 

Questioiis. 

What  is  subtraction? 

What  is  the  greater  number  called  ? 

What  is  the  less  number  called  1 

What  is  the  difference  called  ? 

How  do  you  place  numbers  for  subtraction? 

When  the  lower  figure  is  greater  than  the  upper  one,  how  do  you 
proceed  ? 

Why  is  the  07ie  you  borrow,  one  ten  ? 

Ans.  Because  ten  ones  make  one  ten  ;  and  if  I  borrow  one  ten,  it 
will  make  ten  ones  again,  &c. 

How  do  you  prove  subtraction  ? 


MULTIPLICATION. 

(Art.  11.)     Multiplication  is  the  repetition  of  the  same  num- 
ber, hence,  it  can  be  performed  by  addition. 


MULTIPLICATION. 


25 


Tlie  number  to  be  repeated  is  called  the  multipHcancf,  the  number 
of  times  it  is  to  be  repeated  is  called  the  multiplier,  and  the  result  is 
called  the  product.  'I'he  multiplier  and  multiplicand  are  also  called 
factors  of  the  product. 

EXAMPLE. 

Let  8  be  repeated  three  times,  thus:  8-}-8-f-8=24.  The  short 
mental  operation  of  saying  and  comprehending  tiiat  3  times  8  are  24, 
is  called  multiplication  ;  8  is  called  the  multiplicand,  3  the  multiplier, 
and  24  the  product.  But  this  is  arbitrary  :  we  may  call  3  the  multi- 
plicand, and  8  the  multiplier ;  and  it  will  then  be  required  to  repeat 
3,  8  times :  hence,  in  point  of  fact,  it  is  indifferent  which  of  the  two 
factors  is  called  the  multiplier,  and  which  the  multiplicand. 

Any  two  or  more  numbers  which,  multiplied  together,  produce  a 
third  number,  are  cMed  factors  of  the  number.  A  number  may  have 
many  different  factors ;  for  example,  36  may  have  for  factors  3  and 
12,  2  and  18,  4  and  9,  6  and  6,  or  2,  3  and  6,  because — 

3X12=36 
2X18=36 
4X  9=36 
6X  6=36 
2X3X   6=36. 

MULTIPLICATION    TABLE. 


2X    1=  2 

iX    1=  4 

6X    1=  6 

SX    1  = 

8 

lOX    1=   10 

2X   2=  4 

4X   2=  8 

6X   2=12 

8X   2= 

16 

lOX   2=  20 

2X   3=  6 

iX   3=12 

6X   3=18 

8X   3= 

24 

10  X   3=  30 

2X   4=  8 

4X  4=16 

6X  4=24 

8X  4= 

32 

lOX   4=  40 

2X   5=10 

IX    5=20 

6X   5=30 

SX   5= 

40 

lOX   5=  50 

2X   6=12 

iX   6=24 

6X   6=36 

8X   6= 

48 

lOX   6=  60 

2X   7=14 

4X   7=28 

6X   7=42 

SX   7= 

56 

lOX   7=  70 

2X   8=16 

4X   8=32 

GX   8=48 

8X   8= 

64 

lOX   8=  80 

2X   9=18 

IX   9=36 

.^X   9=54- 

SX   9= 

72 

10  X   9=  90 

2X10=20 

4X10=40 

0X10=60 

8X10= 

80 

10X10=100 

2X11=22 

4X11—11 

SX  11=66 

8X11= 

88 

10X11  =  110 

2X12=24 

4X12=48 

6X12=72 

8X12= 

96 

10X12=120 

3X    1=  3 

5X    1=  5 

7X    1=  7 

9X    1= 

9 

IIX    1=  11 

3X   2=  6 

5X  2=10 

7X   2=14 

9X   2= 

18 

11 X   2=  22 

3X   3=  9 

5X   3=15 

7X   3=21 

9X   3= 

27 

I IX   3=  33 

3X  4=12 

5X  4=20 

7X  4=28 

9X   4= 

36 

11  X  4=  44 

3X   5=15 

3X   5=25 

7X   5=35 

9X   5= 

45 

1 1  X   5=  55 

3X    6=18 

5X   6=30 

7X   6=42 

9X   6= 

54 

IIX   6=  66 

3X    7=21 

ox   7=35 

7X   7=49 

9X   7= 

63 

11  X   7=   /7 

3X   8=24 

5x   8=40 

7X   8=56 

9X   8= 

72 

11 X   8=  88 

3X    9=27 

5x   9=45 

7X   9=63 

9X   9= 

81 

IIX   9=  99 

3X10=30 

5X10=50 

7X10=70 

9X10= 

90 

11X10=110 

3X11=33 

5X11=55 

7X11=77 

9X11  = 

99 

11X11=121 

3X12=36 

5X12=60 

7X12=84 

9X12= 

108 

11X12=133 

26  ARITHMETIC. 


As  multiplication  is  repeating  a  sum  a  given  number  of  times,  it  is 
evident  that  the  units,  and  tens,  and  hundreds,  and  every  superior  or- 
der of  the  proposed  sum,  must  be  repeated  the  given  number  of  times  ; 
and  in  repeating  the  units,  if  they  amount  to  any  number  of  tens,  and 
units,  the  tens  must  be  added  into  the  column  of  tens,  when  we  re- 
peat the  tens,  for  the  same  reason  that  we  carry  by  tens  in  addi- 
tion. 

To  analyze  the  process,  let  it  be  required  to  multiply  468  by  8  ;  that 
is,  repeat  468,  8  times. 

468=400-1-  60-|-  8 

Repeat  each  order  8  times,  8 


3200-|-480-f64 


Or,  in  a  more  condensed  form,  it  stands  thus 
Multiplicand,  468 
Multiplier,  8 


Product,  3744 


From  this  we  perceive,  that  in  all  cases  the  tens  which  arise  from 
repeating  the  units  must  be  added  in  w4th  the  tens  that  arise  from  re- 
peating the  tens;  and  the  hundreds  which  arise  from  repeating  the 
tens,  must  be  added  to  the  column  of  hundreds,  &c. ;  hence,  when 
the  multiplier  is  a  single  Jigure,  we  have  the  following 

RuLT..  Place  ihe  multiplier  under  the  units  figure  of  the  multi- 
plicand, and  mtiltiply  each  figure  cf  the  multiplicand  in  succession, 
and  set  down  the  amount,  and  carry  as  in  addition, 

(AuT.  12.)  Pnoor.  The  proof  of  multiplication  is  division,  for 
they  are  reverse  operations  ;  but  this  method  of  proof  cannot  be  used 
until  the  pupil  has  learned  division.  But  if  we  vary  the  method  of 
multiplying,  and  find  the  same  result,  we  may  be  quite  conlident  that 
the  result  is  correct;  and  some  persons  would  regard  it  as  proof  and 
others  would  not.  For  instance,  if  we  have  any  number,  say  231,  to 
be  multiplied  by  8.  We  may  multiply  it  by  8  directly,  and  then  for 
proof,  divide  8  into  any  two  parts,  say  6  and  2,  or  5  and  3,  and  mul- 
tiply 231  by  6;  then  multiply  231  by  2,  and  the  two  products  added 
together  will  be  the  same  as  the  first  result,  if  no  mistakes  have  been 
made.     Hence,  to  prove  multiplication,  we  have  the  following 

Rult:.  Separate  the  multiplier  into  any  two  parts,  and  multi- 
ply by  such  part  separately,  and  add  the  two  results  together,  and 
their  sum  will  be  the  same  as  by  direct  multiplication,  provided  no 
mistakes  have  been  made. 


MTLTIPLICATION.  27 


EXAMVLES. 

(1)                                       (2)  (3)  (4) 

3G563         8375  4378  9286 

5           6  7  8 


182315        50250        ;30G46        74288 

APPLICATION-. 

1.  If  I  barrel  of  flour  sells  for  555  cents,  what  will  6  barrels  come 
to  "^  Ans>.  33:30  cents. 

2.  If  1  load  of  hay  costs  875  cents,  what  will  7  loails  come  to? 

A?}s.  6125  cents. 

3.  If  1  cord  of  wood  costs  486  cents,  what  will  8  cords  come  to? 

Ans.  3888  cents. 

4.  If  I  yard  of  broadcloth  costs  765  cents,  what  will  9  yards  come 
to  ?  Ans.  6885  cents. 

5.  A  lady  purchased  54  yards  of  muslin  at  12  cents  a  yard  ;  what 
was  the  cost  of  the  whole  1  Am:  648  cents. 

6.  A  farmer  sold  49  acres  of  land  at  U  dollars  an  acre  ;  what  did 
the  whole  amount  to?  Ans.  5.i9  dollnrs. 

7.  A  Carpenter  worked  ten  days  at  146  cents  a  day  ;  ho.v  much  did 
his  wages  amount  to?  Ans.  1460  cents. 

8.  VVliat  will  764  pounds  of  pork  come  to  at  8  cents  a  pound  ? 

Ans.  61 12  cents, 

9.  A  person  spends  9  dollars  a  week  ;  how  much  will  he  spend  in 
4G  week«  ?  Ans.  4  14  dollars. 

10.  If  ynu  save  12  cents  a  week,  how  much  will  you  save  in  52 
weeks,  or  1  year]  Ans.  624  cents. 

11.  James  spends  for  nuts  and  cakes  8  cents  a  day  ;  how  much 
will  he  spend  in  a  year,  or  365  days?  Ans.  2920  cents. 

12.  .lane  spends  for  cakes,  nuts,  and  apples,  6  cer-ts  a  dav  ;  how 
much  will  she  spend  in  216  days?  Ans.  1296  cents. 

(Aht.  13.)  Before  we  proceed  any  further,  it  is  important  to  re- 
mark, that  multipliers  must  always  be  considered  as  abstract  num- 
bers. Mu!tiplicaii(m  is  a  mere  repetition  of  the  Same  number,  or  the 
same  //tin<{;s,  a  given  numl)er  of  times.  'Vhe  number  expressing  the 
times  anything  is  to  be  taken,  must  be  mere  count,  and  cannot  repre- 
sent things. 

In  the  first  example,  if  1  barrel  of  flour  sells  for  555  cents,  vi'hat 
will  6  barrels  come  to? 

Here  the  555  cents  cannot  be  mulfipl'ud  by  barrels,  but  it  can  and 
must  be  repeated  as  many  times  as  there  are  barrels.*  'i'he  same 
reason  applies  to  all  other  questions. 

The  quantity  which  has  the  same  name  as  the  required  result  is 

*Tliis  maybe  considered  a  nice  dislinciion ;  hut  it  is  essential,  as  we  find 
so  much  carelessness  on  this  poini  arnon,£r  some  of  our  teachers  and  in  some 
of  our  hooks.     It  is  not  unfr.-ciuenltlia!  we  find  such  demands  as  the'^e  :  Mul- 


28  ARITHMETIC. 


always,  properly  speaking,  the  multiplicand,  whether  we  practically 
make  it  so  or  not.  Thus,  in  the  12th  example:  Jane  spends  for 
cakes,  nuts,  and  apples,  6  cents  a  day  ;  how  much  will  she  spend  in 
216  days  ?  Here,  6  cents  must  be  taken  216  times,  but  in  practice 
wc  write  the  operation  thus  ; 

216 
6 

1296 

The  largest  number  is  taken  for  the  multiplicand,  for  it  makes  no 
difference  :  the  product  of  two  factors  is  the  same,  whichever  be  re- 
garded as  the  multiplicand. 

When  the  multiplier  consists  of  more  than  one  figure,  the  product 
of  the  superior  figure  will  be  a  supeiior  product  ;  for  instance,  the 
product  of  6  tens,  or  60,  must  be  10  times  as  much  as  the  pro;luct 
of  6  ;  though  we  may  multiply  by  any  superior  figure,  in  the  same 
manner  as  we  do  by  the  unit  figure,  only  taking  care  to  give  the 
product  its  true  place  in  the  07'der  of  numbers.  From  these  consid- 
erations arises  the  following 

Rule.  When  the  multiplier  contains  more  than  one  JJgure, 
multiply  the  multiplicand  by  each  figure  of  the  multiplier,  placing 
the  right-hand  fig^-ire  of  each  product  directly  under  that  figure 
(f  the  multiplier  Ijy  luhic'i  it  is  produced,  and  take  the  .sum  of  all 
the  products. 

EXAMPLES. 

37462 
.563 


Product  of  the  units, lll^SSo 

Product  of   the  tens, 224773 

Product  of  the  hundreds, 187310 


Entire  product  of  563, tilD9ll06 

2.    Multiply   4962  .    .  by  .    .    .  9S Ans.  486276. 

3 7331  ..!...  87 642108. 

4 4376 97 424472. 

5 7923 73 617994. 

6 6842 89 608938. 

7 7648 523 3999904. 

P 8473  .....     456 3S63688, 

9 9372 567 5313924. 

10 75649 579 43800771. 

liply  25  cents  by  25  cents;  multiply  2  shillings  and  6  pence  by  2  sliiM^ng-?  and 
6  pence.  &c.  Now.  we  can  repeal  25  cents,  or  any  other  ^^mn  of  money,  as 
viany  times,  or  an  many  parts  of  one  time  as  we  pll'a^;o :  lun  tliere  is  no 
such  thing  as  money  times  money,  or  imrreis  times  barrels,  or  barrels  times 
money.  If  it  were  possible  to  take  money  times  money,  what  would 
it  produce? 


MULTIPLICATION.  29 


11.    .Multiply  29S31  .    .by.    .     952.    .    .    .^77,^.28399112. 

12 252:J8  .    .    .    .     12170 30714G4G0. 

13 991)9 6tJ(iO 66593.340. 

14 87603 9865  .....      861203o95. 

(AuT.  14.)  To  multiply  by  10,  we  place  a  cipher  at  the  right; 
'i'hus,  5  multiplietl  by  10,  gives  50;  that  is,  the  cipher  is  put  at  ihe 
right  of  the  5.  In  the  same  way,  to  multiply  by  100,  we  put  two 
ciphers  thus— 500;  6  multiplied  by  1000,  wc  have  6000,  &c. 


kxa: 

IPI.E^-. 

10. 

Multiply  47654  . 

.by 

.  10  .  .  . 

.  .  Ans.   476540. 

11. 

.....  75478  . 

.   100  .  . 

.  .  .   7547600 

12. 

86427  . 

1000  .  .  . 

.  .  .  8G427000. 

13. 

17846  . 

10000  .  . 

.  .   178160000 

14. 

lOOCO  . 

.  . 

17S46  .  . 

.  .   178160000 

It  will  be  observed  that  examples  13  an.l  14  are  tlic  same,  and  pro- 
duce the  sajne  results,  ft  i.s  immaterial  whether  the  ciphers  are  in 
one  factor  or  the  other,  or  in  hotk  ;  the  result  will  have  the  same  num- 
ber of  ciphers  on  the  right  as  are  contained  in  both  factors.  Henct, 
rvheji  ficlors  have  ciphers  on  their  right,  the  ciphers  may  be  omit- 
ted in  the  operation,  and  afterwards  annexed  to  the  product.  Thus  : 

15.  Multiply  47000  by  4200  Operation  .• 

47 
42 

9i 

188 

Kesuit,    197400000 


16.     MuUiplv  73400  by  72000 Result,  3082SOOOOO. 

17 '.  62300  bv  12100 751410000. 

18 785400  by     2000 1570800000. 

When  ciphers  are  between  numbers  in  the  multiplier  there 
will  be  no  product  corresponding  to  then),  because  they  signify 
nothin<T  or  770  times.  The  other  partial  products  exist  the  same  as 
before. 

EXAMPLES. 

Operation. 
1.  Multiply  672  by  602.  672 

^  -^  603 


1344 
4032 


404544 


c  2 


30  ARITHMETIC. 


Here  are  two  partial  products  ;  the  product  for  the  units,  and  the 
product  for  the  hundreds.  'J'here  is  no  product  for  the  tens,  because 
the  multiplier  contains  no  tens. 

2.    Multiply    96038  .    .  by  .       6007.  .    .    .  Ai^s.   576900266. 

3 40307  ....     30508 1229685956. 

4 3000024 309 1170007416. 

5 83200 904 75212800. 

GENERAL    APPLICATION    OF    MULTIPLICATION. 

1.  A  gentleman  owes  25  laborers  16  dollars  each  ;  how  many  dol- 
lars will  be  required  to  pay  them  ?  Ans.  400  dollars. 

2.  A  carpenter  owes  a  journeyman  for  43  days'  work,  at  125  cents 
per  day  ;  what  did  the  whole  amount  to  ?  An.s.  5375  cents. 

3.  A  merchant  buys  440  yards  of  muslin  at  17  cents  per  yard  ; 
what  was  the  amount  of  the  bill  1       Am.  7480  cents,  or  $74.80. 

4.  A  farmer  sold  60  bushels  of  wheat  at  125  cents  per  bushel, 
40  bushels  of  rye  at  85  cents  per  bushel,  and  34  bushels  of  corn  at 
50   cents   per   bushel ;   how  much  money  did  he  receive  ? 

,'      .         Ans.  126  dollars. 

5.  A  hogshead  contains  63  gallons  :  what  will  3  hogsheads  of 
molasses  cost,  at  27  cents  per  gallon  ?  Ans.  'Si-51.03. 

6.  A  bushel  contains  32  quarts  :  what  will  4  bushels  of  chestnuts 
come  to,  at  5  cents  per  quart  ?  Ans.  640  cents,  or  $6.40. 

7.  VVliat  will  396  bushels  of  potatoes  come  to  at  24  cents  per 
bushel  ^  Ans.  §95.04. 

8.  How  many  dollars  will  be  required  to  pav  for  29  fat  oxen,  at 
43  dollars  each  1  "'  Ans.  $1247. 

9.  A  certain  wagon  wheel  will  turn  round  350  tinjcs  in  running 
a  mile  :  how  many  times  will  it  turn  round  in  running  25  miles  ? 

Ajis.  8750  times. 

10.  The  semi-diameter  of  the  earth  is  3956  miles,  and  it  is  60 
times  that  distance  to  the  moon;  and  the  distance  to  the  sun  is  400 
times  the  distance  between  the  earth  and  moon ;  how  many  miles  is 
it  10  the  sun?  /i«s.  94944000" miles. 

1 1 .  There  are  360  degrees  in  the  circumference  of  the  earth  ;  and 
if  each  degree  were  60  miles,  what  would  be  the  number  of  miles 
round  the  earth  ?  If  70  miles  to  a  degree,  what  would  be  the  num- 
ber of  miles  ?  a       S^^  60— 21 600  miles. 

^"^' 1  at  70— 25200  miles. 
N.  B. — The  circumference  is  24856  miles, 

12.  If  a  ship  should  sail  8  miles  an  hour  for  18  days,  how  far 
would  it  sail?  Ans.  3456  miles. 

13.  How  many  panes  of  glass  are  there  in  a  house  that  has  20 
windows,  each  containing  24  panes,  and  12  windows,  each  contain- 
ing 18  panes?  Ans.  696  panes. 

14.  There  are  60  minutes  in  an  hour;  how  many  minutes  are 
there  in  24  hours,  or  one  day  ?  ,  Ans.  1440. 

15.  How  many  minutes  are  there  in   365  days? 

Ans.  525600. 


MULTIPLICATION.  31 


16.  A  house  contains  26  windows,  each  containing  24  panes  of 
glass;  how  many  panes  are  there  in  the  whole  ?  Aus.  6:^4. 

17.  There  are  20  shillings  in  a  pound  ;  how  manv  arc  there  in  456 
pounds]  Ans.'dl20  shillings. 

18.  There  are  12  pence  in  1  shilling;  how  many  pence  are  in  254 
shillings  ?  yl??5-.  3048. 

19.  There  are  24  hours  in  1  day;  how  many  hours  are  in  156 
days?  Alls.  3024:. 

20.  There  are  28  pounds  in  1  quarter  ot"  a  cwt. ;  how  many  pounds 
are  in  124  quarters';'  yl/^^■.  3472. 

21.  There  are  20  pieces  of  cloth,  each  containing  37  yards,  and  49 
other  pieces,  each  containing  75  yards ;  how  many  yards  of  cloth 
are  there  in  all  the  pieces?  Ans.  4415. 

22.  Thoro  are  24  hours  in  a  day,  and  7  days  in  a  week  ;  how 
many  hours  in  a  week  ?  Ans.  163. 

23.  A  merchant  buys  a  piece  of  cloth  containing  97  yards,  at  3 
dollars  a  yard  ;  what  does  the  piece  cost  him  1     An^.  291  dollars. 

(Art.  15.)  A  composite  number  is  such  a  number  as  may  be  pro- 
duced by  the  product  of  two  or  more  factors  multiplied  together, 
'i'hus,  24  is  a  composite  number,  because  4  times  6  make  24,  or  2 
times  12  make  24,  or  3  times  8  make  24,  or  2 X 2 X 2  X ;^="-i4  ;  also, 
56  is  a  coniposHe  number,  because  it  can  be  composed  of  the  product 
of  7  times  8.  or  2X^X7,  &c. 

Obsrvation. —  When  a  multiplier  is  a  composite  numuer,  we  may 
mitltiply  the  multiplicand  by  one  factor,  and  that  product  by  an- 
other, and  so  on,  until  all  the  factors  are  used.  The  lad  product 
will    be    the    result    required. 

EXAMPLES, 

1.  Multiply 476  by  24 

As  24=4X6,  multiply  by 4    ' 

1904 
Now,  this  product  by 6 

Result, 11424 

2.  Multiply  425  by  16.     (16=4X4)  Ans.  mOQ. 

3.  Multiply  2873  by  54.    (54=9X6)  Ans.  155142. 

4.  Multiply  108  by  28.      (28=4X7)  An.^.  3024. 

5.  Multiply  2042  by  484.  (484=12  IX  4) 

(121=  IIXH)  (4=2X2) 
2042X11  X  1 1  X2X2=988324, 
Asa  pure  mental  operation  for  one  who  is  blind,  or  has  no  means 
at  hand  of  writing  figures,  this  may  be  preferable  to  the  common 
method  ;  but,  under  ordinary  circumstances,  the  conmion  way  is  the 
best,  for  the  factors  are  co!i:imonly  larger  figures  than  their  aggregate 
product.  For  in.stance,  if  we  were  required  to  multi[)ly  by  63.  the 
factors  of  this  product  are  7  and  9,  larger  numbers  than  the  mere 
figures  6  and  3. 


32  ARITHMETIC. 


There  are  many  other  ways  of  performing-  niuUiplication,  but  we 
cannot  explain  many  of  them  before  we  acquire  the  art  of  dividing. 

Tl)is  principie.  as  last  explained,  however,  is  not  strictly  confined 
to  composite  numbers,  if  the  pupi!  will  consider  the  natuie  of  niulti- 
plicaiion.  the  real  object  sought  fur.  It  is,  to  obiaia  Hit  amount 
i:J'  l/ic  multiplicand  as  many  times  as  there  are  units  in  the 
multiplier. 

Now,  suppose  we  wish  to  multiply  any  number  by  37,  which  is 
not  a  composite  number,  we  may  consider  that  37=36-1-1,  and 
36=6X6.  Therefore,  wo  may  obtain  the  multiplicand  36  times, and 
then  adil  it  once  to  f.)rm  the  product.  In  the  same  manner,  to  multi- 
ply by  29,  we  may  first  multiply  by  30,  and  subtract  once  the  multi- 
plicand, &c.  But  these  subjects  will  come  up  again  in  the  supple- 
ment to  multiplication  and  division. 

Questions. 

Does  it  make  any  difference  in  the  product,  which  of  the  two 
given  numbers  you  make  the  multiplier? 

Why  is  the  largest  number  usually  made  the  multiplicand,  and 
the  smallest  the  multiplier  ? 

Multiplication  is  but  another  method  of  performing  addition  :  is 
it,  then,  indispensable  in  practice? 

Why  do  we  place  the  product  of  the  second  figure  in  the  multi- 
plier one  place  to  the  left  of  the  product  of  the  first  figure  ? 

Why,  in  multiplying,  do  we  begin  with  the  lowest  figure  in  the 
multi]  lier,  rather  than  the  highest? 

Provided  we  place  the  products  for  addition  in  the  order  of  their 
respective  values,  does  it  make  any  dilTerence  in  the  result  which 
figure  of  the  multiplier  we  commence  with  1 

If  you  annex  three  cii>her3  to  the  right  hand  of  a  number,  how 
much  is  its  value  increased  ? 


DIVISION. 

(AnT.  16.)      1.  DivisTov  is  the  art  of  finding  how  na any  times 
one  number  is  contained  in  another. 

2.  Division  is  a  short  method  of  finding    how  many  times  one 
number  can  be  subtracted  from  another. 

3.  Division  is  the  converse  of  multiplication,  and  the  same  princi- 
ples are  to  be  preserved  in  both  multiplication  and  division. 

4.  Division  is  a  concise  method  of  finding  how  many  times  one 
number  is  greater  than  another. 

'J'hese  four  definitions  are  but  different  expressions  to  explain  the 
same  priiici[)le. 

The  three  following  technicalities  are  to  be  particularly  noticed  : 

1.  The  dividend,  or  number  to  be  divided. 

2.  The  divisor,  or  number  by  which  it  is  divided. 


DIVISION.  33 


;5.  The  quoiienf,  or  result  of  the  division,  showing  how  many 
times  the  ilivisor  is  contained  in  the  dividend. 

When  tlie  divisor  is  not  contained  in  the  dividend  an  exact  num- 
ber of  times,  what  is  over  or  reinaininir  after  tlic  division,  is  called 
the  remtdnder  ;   it  must,  of  course,  he  less  than   the  divisor. 

The  sign  comrnordy  used  to  denote  division,  is  -i-:  tiius.  r2H-3=4. 
Division  is  also  indicated  by  placing  the  divisor  under  the  dividend, 

n 

with  a  line  between  them:  thus,  — =4,  which  is  read,  VZ  divided  bv 

3 
;3  equals  4,  * 

'I'he  following  illustration  will  better  chicidate  the  subject :  3  can 
be  subtracted  from  12  four  times;  thus, 
12 
3  one  time. 

9 

3  two  times. 

6 

%  3  three  times. 

3 

3  four  times. 

0  remains. 

How  many  hats,  at  3  dollars  apiece,  can  be  bought  for  12  dollars  ? 

How  many  times  can  3  be  subtracted  from  12  1 

How  many  times  can  4  be  subtracted  from  24  ] 

How  many  times  can  8  be  subtracted  from  27  ? 

At  5  miles  per  hour,  how  many  hours  will  it  require  to  travel  40 
miles  ? 

At  4  dollars  a  yard  for  cloth,  how  many  yards  can  be  bought  for 
42  dollars? 

How  many  times  can  4  be  subtracted  from  42,  and  how  many 
will  be  left  after  the  last  subtraction  ? 

The  foregoing  questions  illustrate  the  general  principles  of  divi- 
sion ;  and  the  pupil  will  at  once  perceive  its  connection  with  subtrac- 
tion ;  but  we  can  neither  add  nor  subtract  unlike  quantities  (se(! 
Art.  5).  So  in  division:  the  divisor,  dividend^  and  qiwlient,  wiW 
have  a  philosophical  relation  with  each  other  in  regard  to  things. 

Thus :  If  the  dividend  be  a  concrete  number,  such  as  dollars, 
yards,  miles,  &c.,  and  the  divisor  an  abstract  number,  or  a  mere  nu- 
meral, the  quotient  will  be  of  the  same  name  as  the  dividend. 

If  both  the  divisor  and  the  dividend  are  concrete  munbers  of  the 
same  kind,  the  quotient  will  be  an  abstract  number,  or  a  mere  nume- 
ral.    One  or  the  other,  the  divisor  or  the  quotient,  must  be  a  mere 


34 


ARITHMETIC. 


numeral.     When  hoth  the  divisor  and  the  quotient  are  abstract,  tlie 
dividend  must  also  be  abstract. 

As  division  is  the  converse  of  multiplication,  we  now  proceed 
to  fexplain  it,  by  taking  an  example  in  multiplication  to  trace  its  con- 
verse operation  :  thus, 


Multiply  704  by  4  ;  that  is- 


Multiply 


700-1-60-1-4  by  4, 
4 


2800-1-240-1- lG=r.3056 


The  converse  operation  is,  to  divide  SDfjG  Jjy  4 ;  and  we  must 
make  the  same  division  of  this  number  as  atK'V?,  or.  rather,  the  ope- 
ration itself  will  thus  separate  the  number.  Au^  as  division  must  be 
in  all  respects  the  converse  of  multiplication,  we  rtt  ist  commence  di- 
vision at  the  left,  or  at  the  highest  order  of  the  n«u«ioer.  'J  bus,  4  is 
not  contained  in  3,  (8000,)  and  the  greatest  numlxTT  ^\i:vkv  .30,  vvhith 
is  exactly  divisible  by  4.  is  23,  or  in  this  example,  20^0  ^-r  same 
number  that  we  perceive  in  the  multiplication — and  tlas  e)oaM>ntary 
operation  will  stand  as  follows  :  * 


4)3056(700         700-^ 

28  I 


60  ^Partial  quotient* 


4)256(60 
240 


4)16(4  4  J 

16  764     whole  quotient. 


Divide  3496  by  16, 

Here  we  must  consider  that  16  is  contained  in  34  2  times;  ^W  as 
34  is  34  hundred,  the  2  is  2  hundred,  and  the  operation  is  thus. 


16)3496(200 
3200 


200^ 


16)296(10 
160 


10 


S-Partial  quotients. 


10)136(8 


8  218  whole  quo.  and  rem.  of  8. 

But,  in  place  of  making  three  problems  of  this,  we  can  make  one, 


DIVISION. 


35 


16)3496(218 
32 

29 
18 

136 
128 


From  the    foregoing,  we  draw    the   following 

RcLK.  [•"ind  how  many  times  the  divisor  is  contained  in  as  many 
of  the  kfl-hand  figures  of  the  dividend  as  are  ju^t  sufficient  to  con- 
tain it,  and  set  the  result  on  the  right  of  the  dividend  fur  the  first 
quotient  fgure  ,•  multiply  the  divisor  by  it,  and  place  the  product 
under  the  part  of  the  dividend  taken;  subtract  it  therefrom,  and 
to  the  right  of  the  remainder  bring  down  the  next  figure  in  the 
dividend:  divide  this  number  as  before,  and  thus  proceed  until 
every  figure  in  the  dividend  is  brought  down. 

[NoTK.  Remainders  may  either  be  carried  out  to  the  riglit  of  the 
quotient,  with  a  hue  between,  or  be  placed  above  the  hue,  and  the 
divisor  below,  in  the  form  of  a  fraction.] 

When  the  divisor  does  not  exceed  12,  the  multiplication  and  sub- 
iraciion  may  be  made  in  the  mind,  without  the  formality  offsetting 
down  the  whole  process.  In  this  case,  the  quvtient  is  placed  under 
the  dividend,  with  a  line  drawn  between. 


EXAMPLES. 

(1) 

Divisor  4)654324 

(2) 
6)8574684 

!3) 
8)146237456 

Quotient    163581 

1429114 

18'^  79682 

(4> 
Divisor  10)1786940 

'5) 
9)7468542 

11)564521705 

Quotient       178694 

(7) 
12)9476521764 

(8) 
11)546215747 

(9) 
12)462164684 

1 

36  ARITHMETIC. 

The  operations  may  be  proved  by  multiplying  the  quotients  by  the 
divisor,  and  adding  the  remainder,  if  there  is  any,  to  the  product.  If 
the  v»^ork  is  correct,  it  will  equal  the  dividend. 

10.  What  will  be  tlie  quotient  of  4562,  divided  by  3  ] 

Quo.  1520,  Rem.  2. 

11.  What  will  be  the  quotient  of  44631,  divided  by  4? 

Quo.  11157,  Teem.  3. 

12.  What  will  be  the  quotient  of  62070,  divided  by  4? 

Quo.  155)7,  Rem.  2. 

13.  What  will  be  the  quotient  of  5374,  divided  by  5  ? 

Quo.  1074,  Rem.  4. 

14.  What  will  be  the  quotient  of  7856,  divided  by  6  1 

Quo.  1309,  Rem.  2. 

15.  What  will  be  the  quotient  of  85362,  divided  by  7  ? 

Quo.  12194,  Rem.  4:. 

16.  Divide  ....  958  by  .      18.  .    .    .  Quo.    53    Rem.  4. 

17.  Divide.    .    .     1475  by.      28 .52  .    .  .  19. 

18.  Divide  .    .    .    4277  by  .      31 137  ..  .  30. 

19.  Divide  .    .    .  25757  by  .      37 696  ..  .  5. 

20.  Divide.    .    .63125  by.    123..   •     .    .    .    513  .    .  .  26. 

21.  Divide  .    .     253622  by  .    422 601  ..  . 

22.  Divide  .    .  4049160  by  .    328 12345  .    .  . 

23.  Divide  .    48905952  by    9876 4952  .    .  . 

24.  Divide.           25312  by  .    226 112  ..  . 

25.  Dividj  .    .    .  43956  by  .    666 66  .    .  . 

26.  Divid-  .  101442075  by    4025 25203.    .  . 

27.  Divide'  .  367376610  by  158694 2315.    .  . 

(Art.  17)  As  division  is  the  converse  of  multiplication,  and  as 
we  niiiltinly  by  10,  100,  1000,  &c.,  by  annexing  the  ciphers,  (Anx. 
14.)  thcrctore,  when  we  cut  off  ciphers,  we  divide  by  10,  100,  &c., 
according  to  the  number  of  ciphers  cut  off.  But,  if  there  are  no 
ciphers  to  be  cut  off.  and  we  wish  to  divide  by  10,  100,  1000,  &;c., 
we  may  divide  by  the  following 

RiTLE.  Cut  off  as  many  figures  from  the  right  of  the  dividend 
as  there  are  ciphers  in  the  divisor.  The  figures  on  the  left-hand  of 
the  point  will  be  the  quotient,  and  those  on  the  right  hand 
the  remainder. 

r  10     =785.4  . Ans.  785_4_ 

7854^^   100  =78.54 78^_1 

1  10  0 

/  1000=7.854 7J_5_4_ 

^  10  0  0 

Multiplying  a  number  by  10  increases  its  value  tenfold  ;  and  di- 
viding it  by  10  ditninishes  it  in  the  same  ratio.  The  former  is  effect- 
ed by  removing  every  figure  in  it  one  place  farther  from  the  unit's 
place  ;  the  latter,  by  bringing  them  one  place  nearer. 

When  there  are  ciphers  on  the  right  hand  of  the  divisor : 


DIVISION.  37 


Cut  off  lite  ciphers,  and  an  equal  number  of  places  from  the 
rii^ht  of  the  dividend:  in  dividinii;,  omit  the  fih;urcs  cut  off,  and 
annex'  tJicm  afterwards  to  the  remainder,  if  any  ;  othtrwiac,  the 
fiirin-cs  cut  off  from  the  dividend  are  the  remainder. 


Divide  3704196  by  20.  Divide  369183  by  7100. 

(1;  (2) 
2,0)3704196  7i;00)3r)91|83(51T  o.|3  ^,4ns. 
'       355 


Ana.   18520916 

2  0 


141 
71 


70 


3.  Divide  2976435  by  2800. 

4.  Divide  9400639  by  4700. 
.5.  Divide  6749802  by  9000. 
6.  Divide  4872036  by  1200. 

Wben  the  dividend  is  federal  money,  the  operation  is  the  same  as 
in  division  of  whole  numbers,  and  the  only  difficulty  is  to  decide  the 
value  of  the  quotient ;  but  this  will  be  treated  more  at  large  when  we 
come  to  division  in  decimal  fractions. 

EXAMPLES. 

!.  Divide  $375.50  equally  among  45  persons. 

Call  the  whole  sum  37550  cents,  and  divide  by  45. 

cents,     cents. 
45)37550(8342  <L;  or,  8  dollars  34^  cents. 

360  *^ 

155 
135 


200 
180 

20 

e  c. 

2.  Divide    56  15  by     10 Quotient,  $  5.61^ 

3.  Divide    96.00  by      5 19.20 

4.  Divide  156  00  by      4 39.00 

5.  Divi(fe    58.14  by     38 1.53 

6.  Divide  417.96  by  129 3.34 

7.  Divide  494.45  by  341 J.45 

8.  Divide  627.38  by  508 1.25^ 


D 


38  ARITHMETIC. 


APPLICATION. 

1.  Suppose  2072  trees  planted  in  14  rows,  how  many' trees  will 
there  be  in  each  row?  Ans.  148. 

2.  Several  boys,  who  wont  to  gather  nuts,  collected  4741,  of  which 
each  boy  received  431  ;  how  many  boys  were  there  1  Aufi.  1 1. 

3.  If  the  ex})ense  of  erecting  a  bridge,  which  is  15036  dollars,  be 
equally  defrayed  by  179  persons,  what  must  each  pay  ? 

Ans.  84  dollars. 

4.  Tt  is  computed  that  the  distance  to  the  sun  is  95,000,000  miles, 
and  that  light  is  8  minutes  traveling  from  the  sun  to  the  earth  ;  how 
many  miles  does  it  travel  per  minute?  Ans.  11875000. 

5.1  A  manufacturer  paid  289  dollars  a  week  lo  his  laborers,  giving 
on  an  average,  62i  cents  per  day  ;  how  many  laborers  had  he  ? 

A71S.  25. 

6.  A  person  bought  27  yards  of  cloth,  for  which  he  paid  6750 
cents;  how  much  did  he  pay  per  yard?  Ans.  250  cents. 

7.  There  are  7  days  in  a  week  ;  how  many  weeks  in  a  year  of  365 
days  ?  Ans.  52  weeks,  and  1  day  over. 

8.  There  are  24  hours  in  a  day ;  how  many  days  in  2040  hours  ? 

Ans.  85  days. 

9.  Twenty-three  persons  dined  together  :  their  bill  was  46  dollars  ; 
how  much  had  each  one  to  pay  ]  Ans.  2  dollars. 

10.  The  sum  of  5500  dollars  was  distributed  in  shares  of  44  dollars 
each;  how  many  shares  did  the  sum  contain?  Ans.  125, 

11.  A  person  spent  2880  dollars  for  land,  paying  24  dollars  per 
acre;  how  many  acres  did  he  purchase  ?  Ans.  120. 

N.  B.  In  resolving  this  question,  we  divide  dollars  by  dollars, 
which  some  have  said  could  not  be  done,  because  we  cannot  multiply 
money  by  money,  as  explained  in  Art.  12  ;  but  the  difficulty  vanishes 
when  we  consider  the  quotient,  120,  as  an  abstract  number,  or  a  mere 
numeral,  as  it  reallij  is.  In  respect  to  the  aritiiinetical  operation,  this 
120  is  no  more  acres  of  land  than  it  is  yards  of  cloth,  or  barrels  of 
flour,  or  anything  else.  The  120  is  merely  the  number  of  times  that 
24  dollars  can  be  subtracted  from  2SS0  dollars  ;  and  in  respect  to  this 
question  we  know,  (not  by  arithmetic,  hut  by  logic.)  that  the  number 
of  acres  miist  lie  the  same  as  this  abstract  number.  We  repeat  once 
more  that,  in  division,  one  or  the  other,  the  divisor  or  the  quofieTif, 
must  be  abstract;  and  this  is  sufficient  to  correspond  to  Art.  12,  in 
multiplication. 

12.  If  we  have  24  baskets  to  pack  1560  eggs,  how  many  must  be 
put  into  a  basket  ?  Ans.  65. 

13.  If  459S0  pounds  of  bread  are  to  be  distributed  among  2120 
soldiers,  how  many  pounds  shall  each  one  receive?  Ans.  19. 

14.  How  many  times  can  a  three-pint  measure  be  filled  from  a 
barrel  of  beer,  containing  36  gallons  ?  J^ns.  96  times. 

We^iroduce  problem  14,  (which  rather  belongs  to  reduction.)  to 
enforce  one  important  principle  in  relation  to  division,  which  is,  that 


DIVISION.  39 


whan  the  tiividcnd  and  divisor  are  holh  coxchki-k  numbers,  tlu  v  must 
both  refer  to  the  name  thing,  and  Oe  of  the  same  drivinniia- 
tion,  because  division  is  but.  a  concise  method  of  performing  certain 
subtractions,  and  we  cannot  subtract  unlike  things.  For  instance,  they 
must  both  be  dollars,  (as  in  example  I !,)  or  both  yards,  &c.  They 
must  both  1)0  of  the  same  denomination ;  one  cannot  be  dollars  and 
the  other  cents  ;  one  cannot  be  yards  and  the  other  miles,  &c.  So  in 
example  14  :  we  cannot  divide  (ofi)  gallons  by  (3)  pints,  as,  at  first 
view,  the  question  seems  to  require ;  and  before  we  can  divide,  we 
must  find  the  number  of  pints  in  30  gallons,  and  divide  that  number 
by  3.  That  is:  When  the  divisor  and  dividend  are  both  concrete 
numbers  they  must  he  of  the  same  dcnotnination.'* 

The  example  is  solved  thus  :  One  gallon  contains  8  pintf?,  therefore, 
multiply  36  l)y  8.  and  the  dividend  will  be  pints,  which  divide  by  3 
pints,  and  we  have  06  for  the  times  the  small  measure  can  be  filled. 

15.  Into  how  many  parts  must  I  divide  the  number  8164,  so  that 
each  part  shall  be  27,  and  leave  a  remainder  of  10  7         Ans.  302. 

!&.<.  A,  B,  and  C,  engaged  to  do  a  job  of  work  for  228  dollars,  and, 
together,  they  accomplished  it  in  40  days.  Tf ow  it  was  agreed  that 
A  should  have  10  cents  a  day  more  than  B,  and  B  10  cents  a  day 
more  than  C ;  what  was  each  man's  share  ? 

Ans.  A  $80,  B  $76,  and  C  $73. 

N.  B.  There  will  be  no  difficulty  in  this  problem,  provided  the 
extra  portions  to  A  and  B  are  taken  from  the  whole  sum,  before 
division . 

Questions. 

What   is  division? 

What  do  you   call    the   number   that   is   to  be   divided  1 

What  do  you  call  the  number  you  divide  by  ? 

What  do  you  call  the  number  obtained  by  division  ? 

What  do  you  call  that  which  is  left  when  the  work  is  done  ? 

When  the  divisor  does  not  exceed  12,  how  do  you  perform 
the  operation  ? 

When  the  divisor  exceeds  12,  how  do  you  proceed  1 

How  do  you  prove  division  ? 

How  may  the  operation  be  performed  when  there  are  ciphers  at  the 
right  hand  of  the  divisor  ? 

When  the  dividend  and  quotient  are  concrete  numbers,  that  is,  re- 
fer to  yards,  pounds,  or  to  any  other  special  thing,  what  will  be  the 
character  of  the  divisor  ? 

Can  we  divide  dollars  by  dollars'!  and  if  so,  what  will  the  quo- 
tient be  ] 

V.AW  we  divide  dollars  by  cents?  if  so,  why  ?  if  not,  why? 

When  the  dividend  is  abstract,  what  must  be  the  character  of  the 
divisor  anJ  quotient  ? 

Can  we  divide  bushels  by  yards?  if  not,  why? 

*  We  are  thus  particular,  in  hopes  to  induce  per«on5  really  to  study  aritli- 
melic,  and  not  waste  so  much  time  in  working  over  it,  apparently  ex- 
pecting the  fciiovvledge  to  come  to  thera. 


y 


40  ARITHMETIC. 


SECTION     II. 


Orskuvatiox.  We  have  now  gone  over  the  ground-rule-  of  arith- 
metic, as  far  as  the  mere  operations  on  numbers  are  concerned  ;  and 
those  persons  who  are  quick  in  what  precedes,  may  very  pro})er!y  be 
said  to  be  quick  at  figures,  though  some  such  persons  may  be  very 
unsuccessful  in  future  progress;  as  that  will  depend  upon  a  pJdluao- 
pkical,  rather  than  a  numerical  turn  of  mind.*  A  good  reasoner  can 
always  be  a  good  arithmetician  ;  on  the  contrary,  one  may  add,  sub- 
tract, multiply,  and  divide,  with  the  rapidity  of  intuition,  and,  if  not 
a  quick  and  sound  reasoner,  quickness  of  operation  will  only  make 
weak  logic  the  more  glaring.  It  is  a  mistake  to  suppose  that  long 
practice  is  most  essential  to  make  a  good  arithmetician. 

Obskuvatiox  Secoxd.  To  apply  the  knowledge  we  have  already 
acquired  to  the  common-place  business  of  every-day  life,  it  is  neces- 
sary that  we  should  know  something  about  fractions,  although,  as  yet, 
we  are  not  prepared  to  go  into  the  subject  of  fractions  in  full. 

We  only  propose  to  make  use  of  a  few  common  fractions,  such  as 
if  4>  h  ■§»  <^^'3  i'^  ^^^6  following 

SUPPLEMENT     TO 

MULTIPLICATION  AND   DIVISION. 

(AnT.  18.)  To  multiply  by  one-half,  is  to  take  the  multiplicand 
one-half  of  one  iime ,-  that  is,  take  one  half  of  it,  or  divitle  it  by  2. 

To  multiply  by  j,  take  a  third  of  the  multiplicand,  that  is,  divide 
it  by  3. 

To  multiply  by  -f ,  take  §  first,  and  multiply  that  by  2  ;  or,  multiply 
by  2  first,  and  divide  the  product  by  3.-{- 

EXAMPLES. 

1.    What  will  360  barrels  of  flour  come  to,  at  5:J-  dollars  a  barrel. 
At  1  dollar  a  barrel  it  would  be  360  dollars  ;  at  5;^  dollars,  it  would 
be  b\  times  as  much. 


*It  IS  not  generally  understood  that  the  power  of  reasoning  and  the  power 
of  coinpaiatiou  are  separate  and  distinct  faculties;  and.  to  be  a  good  mathe- 
matician, a  person  must  possess  both;  but  it  is  most  important  iliat  the  rea- 
soning power  should  be  the  most  iirominent.  Indeed,  to  become  a  skillfi.ii 
mathematician,  it  is  almost  necessary  that  the  individual  sliould  make  nu- 
merical compulations  with  difficulty  ;  otherwise,  he  may  not  be  sufficiently 
prompted  to  seek  lor  fibbreviatiuus  and  artifices,  wiiich,  more  than  anything 
eisi.'.  give  beauty  and  polish  to  science. 

t  Sometimes  one  operation  is  preferable,  and  sometimes  the  other ;  good 
judgment  alone  can  decide,  when  the  case  is  before  us. 


MULTIPLICATION    AND    DIVISION.  41 


360 


5  times,  .    .    .    1800 
i  of  a  time,  .        90 

Ans $1890 

How  much  at  5|  dollars  per  barrel  ? 

360                                      4)360 
5|  

90=i  of  a  time. 

5  times  1800  3 

270  

270=1  of  1  time. 

§2070  

2.  Forty-eight  men  were  to  receive  $5|  a-piece  ;  how  many  dollars 
were  paiiftlieni  ? 

3.  What  will  15  tons  of  hay  come  to,  at  $7^  a  ton  1 

4.  A  farm  of  156  acres  was  sold  for  $34f  an  acre;  how  much  did 
it  come  to  ? 

5.  At  16-§  cents  a  bushel,  what  will  75  bushels  of  potatoes  cost  ? 

Ans.  1250  cents. 

6.  What  will  27|  yards  of  muslin  cost,  at  9  cents  a  yard  ? 

A71S.  249|  cents. 

7.  What  will  1|  yards  of  cloth  cost  at  225  cents  per  yard  ? 

Aiis.  393^  cents. 

8.  What  is   611  times  35  ?  Ans.  242_'_. 

9.  What  is  350  "times  15i  ?  Ans.  5337^. 

(Art.  19.)  Before  we  attempt  to  divide  by  a  mixed  number,  such 
as  2^,  3i,  5§,  &c.,  we  must  explain,  or  rather  observe  the  principle 
of  division,  namely  :  That,  the  quofient  iviU  be  the  same  if  we  mul- 
iiplif  the  dividend  and  divisor  by  the  same  number.  Thus,  24  di- 
vided by  8  gives  3  for  a  quotient.  Now,  if  we  double  24  and  8,  or 
multiply  them  by  any  number  whatever,  and  then  divide,  we  shall 
still  have  3  for  a  quotient.     16)48(3;    .32)96(3,  &c. 

Now,  suppose  we  have  22  to  be  divided  by  5^  ;  we  may  double 
both  these  numbers,  and  thus  be  clear  of  the  fraction,  and  have  the 
same  quotient.     5^)22(4  is  the  same  as  11)44(4. 

How  many  times  is  1^  contained  in  12  ?  Ans.  Just  as  many  times 
as  5  is  contained  in  48.  The  5  is  4  times  1|-,  and  48  is  4  times  12. 
From  these  observations,  we  draw  the  following  rule  for  dividing  by 
a  mixed  number  : 

Rule.  Multiply  the  whole  number  by  the  lower  term  of  the 
fraction ,-  add  the  upper  term  to  the  product  for  a  divisor ,-  then 
multiply  the  dividend  by  the  lower  term  of  the  fraction,  and  then 
divide. 

S5  ~" 


42  ARITHMETIC. 


EXAMPLES. 

1 .  There  being  5^  yards  in  a  rod,  how  many  rods  in  682  yards  ? 

.5i)682(  Double  both,  11)1364(124  ^/js. 

2.  How  many  barrels  of  flour,  at  §5|  per  barrel,  can  be  bought 
with  $500?  Ans.  Se^, 

3.  Thirty-one  and  a  half  gallons  to  the  barrel,  how  many  barrels 
in  485  gallons?  Ans.  \.5'll, 

4.  How  many  times  is  li  contained  in  36?  A?is.  30  times. 

N.  B.  If  we  multiply  both  these  numbers  by  5,  they  will  have 
the  same  relation  as  before,  and  a  quotient  is  nothing  but  a  relation 
between  two  numbers.  After  multiplication,  tlie  numbers  may  be  con- 
sidered as  having  the  denominaiion  of  fiflhs. 

5.  How  many  times  is  ^  contained  in  12?  Ans.  48  times. 

One-fourth,  multiplied  by  4,  gives  1  ;  12,  multiplied  by  4*  gives  48. 
Now,  I  in  48  is  contained  48  times. 

6.  Divide  132  by  2|.  Ans.  48. 

7.  Divide  121  by  15-^-.  Ans.  8. 

8.  How  many  times  is  ^  contained  in  3?  Ans.  4  times. 

(  Art.  20.)  Operalions  in  arithmetic  may  be  considerably  short- 
ened, by  a  little  attention  to  the  relation  of  numbers.  A  few  exam- 
ples are  adduced  of  contractions  in  multiplication. 

Suppose  it  were  required  to  multiply  a  number — say  746382,  by 
999.  (1) 

Multiply  by  1000  (Art.  13.) 746382000 

Subtract  the  multiplicand, 746382 

Product 745635618 


Here,  it  is  evident  that  the  product  of  1000  exceeds  the  product  of 
999,  by  once  the  multiplicand. 

2.  Multiply  4532  3.  Multiply  4532 

bv  ...    639  by  .    .    .    963 


63=9X7       40788  40788 

285516  285516 


Product  2895948  Product  4364316 


In  both  the  foregoing  examples  we  multiply  the  product  of  9  by  7, 
because  7  times  9  are  equal  to  63. 


MULTIPLICATION    AND    DIVISION. 


43 


Because  9  is  in  the  place  of  hundreds  in  example  3,  the  product  for 
the  other  two  figures  is  stt  two  places  towards  the  right. 

In  this  last  example  we  may  commence  with  the  3  units  in  the  usual 
way  ;  then  that  product  hy  2,  because  2  times  3  arc  6  ;  then  the  pro- 
duct of  3  hy  3,  which  will  give  the  same  as  the  multiplicand  by  9. 
The  appearmice  of  the  work,  would  then  be  the  same  as  by  the  usual 
method,  but  would  be  easier,  as  we  actually  multiply  by  smaller 
numbers. 


4.  Multiply 
by.    .    . 


.    .  40788 
497 

285516 
1993612 

20271636 


Product  of  the  7  units. 
As  7X7=49,  multiply  the  pro- 
duct of  7  by  7. 


Observe,  that  in  this  last  example,  497  is  3  less  than  500,  and  500 
is  ^  of  1000.  Hence,  to  obtain  the  product,  we  take  40788  one 
thousand  times,  which  is  40788000, 

2)40788000 

500  times  is 20394000 

3  times  is 122364 

Product, 20271636 


5.  Multiply  4962,  or  any  other  number,  by  98  ;  that  is,  take  it  98 
times.     If  we  take  it  100  times,  we  have 

496200 
Subtract  2  times  4962,  which  is 9924 


Product, 486276 


6.  Multiply  299X299X299. 
Take  299  300  times,  and  subtract  299,  &c. 

7.  Multiply  999X999X999. 


Product  26730899. 


Prod.  997002999. 


;Art.  21.)     To  multiply  by  25,  take  :J-  of  100  times, 
ro  multiply  by  50,  take  ^  of  100  times. 

'-  by  75,  take  |  of  100  times. 

bv  125,  take  g  of  1000  times. 

by  135,  take  g  of  1000  times-f-10  times. 

by  150,  take  100  times-j-^  of  100  times,  &c.,  &c.,  for 


any  other  aliquot  part  of  10,  100,  or  1000. 


44                                                 ARITHMETIC. 

8.  Multiply  86416  by  135. 

125  times  is 

8)86416000 
.  10802000 

10  times  is 

8641G0 

Product, 

.     11666160 

9.  Multiply  any  number,  say  2636,  by  248 

2636 
248 

Commence  with  the  hundreds,  .    .    . 

Double  this  for  the  lens, 

Double  this  last  for  the  units,  .    .    .    . 

.    .     5272 
.    .     10544 
.    .      210S8 

Product, 

.    .     653728 

Take  the  same  example. 

Multiply  by  the  units: 
3  times  8=24  ;  hence 
21088 
3 

Pro 

2636 
248 

21088 
63264 

63264 

duct,  653728 

Again,  if  we  add  2  to  248,  we  h 
hence, 

250  times  is 

ave  250 

:    250  is  i  of    1000; 
4)2636000 
659000 

2  times  subtracted 

5272 

Product, ,    -    -    - 

.    .     653728 

10.  Multiply  87603  by  9865.     B. 
would  be  a  tedious  operation  ;  but,  le 
—135;  135isl25-[-10;  125  is  ^  o 

876030000                            Subt 
1182G405 

^  the  con 
t  us  obs€ 
f  1000. 

ract  i  : 
10  tim 

But  987 

imon  formal  rule,  this 
rve  that  9865  is  10000 
Therefore, 

8)87603000 

10950375 
eg,      876030 

864203595  Product. 

11826405 

11.  Multiply  818327  by  9874. 
Therefore, 

4  is   10000,  less   126. 

MULTIPLICATION    AND    DIVISION.  45 


8183270000  Less  ^  of  818327000 

Subtract.    .    J 03 109202  

102290S75 


8080160798  Product.  818327 


103109202 


12.  Multiply  188  by  135,  and  that  product  by  15.  Instead  of  do- 
ing this  literally  and  mechanically,  according  to  rule,  we  may  half  188 
twice,  and  double  each  of  the  factors  that  end  in  5,  and  we  shall  have 
47  .  270  .    .  30;   or, 

47 
8  100 


4  70  0 
3  7  6 


Product,  3  8  0  7  0  0 


U  i 


13.  Multiply  anv  numl-er,  say  468232,  by   126">       g 

14. by  246  I      -^ 

15. by  426  ' 

16. by  486 

17. by  612 

18.  Multiply  61524  by  273  and  by  327. 

19.  Multiply  342516  by  7209  and  by  9072. 

20.  Multiply  3764  by  199. 

Take  3764  200  times,  and  from  that  product  subtract  3764. 

21.  Muhiply  764  by  498^. 

Take  764  500  times,  and  from  that  product  subtract  1^  times  764. 

22.  Multiply  396  by  21^,  or,  (which  is  the  same,)  99X87=8700 
—86=8613. 

N.  B.    Ninety-nine  is  i  of  396,  and  87  is  4  times  2 If. 

23.  A  man  spends  99  cents  a  day  for  19  years,  each  year  consisting 
of  365  days  ;  what  is  the  whole  amount  1 

We  may  divide  by  5,  25,  50,  125,  and  other  aliquot  parts  of  10, 
100,  1000,  &c.,  in  the  following  manner: 

How  many  times  is  5  contained  in  20  !  Ans.  The  same  number 
of  times  as  the  double  of  5,  or  10,  is  contained  in  the  double  of  20, 
or  40. 

How  many  times  is  50  contained  in  762?  Ans.  The  same  as  100 
in   1524. 


46  ARITHMETIC. 


How  many  times  is  125  contained  in  21251 
Same  as  250  in  4250; 
Same  as     25  in     425 ; 
Same  as     50  in     850; 
Same  as       5  in       85  ; 
Same  as     10  in     170;  that  is,  17  times. 
The  object  of  these  changes  is,  to  give  the  learner  an  accurate  and 
complete  knowledge  of  numbers,  and  of  division  ;  and  the  result  is 
not  the  only  object  sought  for,  as  many  young  learners  suppose. 
How  many  times  is  75  contained  in  575  ?  or,  divide  575  by  75. 

Am.  7^. 

Divide  800  by   12^ Quotient,  64. 

Divide  27  by  \^ Quo.  1//-,  or  l^i. 

A  person  spent  6  dollars  for  oranges,  at  6^  cents  apiece  ;  how  many 
did  he  purchase  ?  Ans.  96. 

(Art.  22.)  When  two  or  more  numbers  are  to  be  multiplied  to- 
gether, and  one  or  more  of  them  having  a  cipher  on  the  right,  as  24 
by  20,  we  may  take  the  cipher  from  one  number,  and  annex  it  to 
the  other,  without  affecting  the  product;  thus,  24X20,  is  the  same  as 
240X2;  286X1300=28600X13;  and  350X70X40=35X7X4 
XIOOO,  &c. 

Every  fact  of  this  kind,  though  extremely  simple,  will  be  very  use- 
ful to  those  who  wish  to  be  skillful  in  operation. 

(Art.  23.)  Before  we  combine  division  with  multiplication,  let 
us  take  a  more  systematic  view  of  numbers. 

The  following  are  called  prime  numbers,  because  no  one  can  be 
divided  by  any  number  less  than  itself  without  producing  a  fraction : 
1    2   3.5.7.    .    .11.13.    .    .  17  .  19  .    .    .23 29 

31      ....  37  ..    .41  .  43  ...  47 53  ...    . 

*  59  .  61 67  .    .    .  71  .  73 79  .    .    .  83  .    . 

...  89 97  .    .    .  101. 

The  points  represent  the  composite  numbers ;  and  here  it  can  be 
observed,  that  there  are  27  prime  numbers,  and.  of  course,  73  com- 
posite numbers  in  the  first  hundred,  the  prime  numbers  becoming  fewer 
as  the  numbers  rise  higher.     Observe  the  following  series : 

5     10     15     20     25     30     35     40     45     60     55     60, 

and  so  on.  Every  body  knows  that  our  arithmetical  scale  of  numbers 
is  1,  10,  100,  1000,  &c.  Now,  we  wish  the  student  to  observe  the 
numbers, 

5     20     25  ^  50     75     125     500, 

as  being  not  only  in  the  preceding  series,  but  aliquot  parts  of  some 
number  in  our  arithmetical  scale.  For  example,  25  is  i  of  100  ;  125 
is  i  of  1000,  &c. 

We  now  charge  the  student  to  make  his  eye  familiar  with  all  the 
preceding  series — the  prime  numbers  as  being  unmanageable  and  in- 


MULTIPLICATION    AND    DIVISION.  47 


convenient,  and  the  others  the  very  reverse  ;  hut  the  full  importance 
of  such  a  study  cun  only  appear  in  the  .sequel. 

(Art.  24.)  When  it  liecomes  necessary  to  multiply  two  or  more 
numbers  together,  and  divide  hy  a  third,  or  i)y  a  product  of  a  third  and 
fourth,  it  must  he  lifenilli/  done,  if  the  numbers  are  prime. 

For  example:   Multiply  19  by  13,  and  divide  that  product  by  7. 

'I'his  must  be  done  at  full  length,  because  the  numbers  are  prime  : 
and  in  all  such  cases  there  will  result  a  fraction. 

But,  when  two  or  more  of  the  numl)ers  are  composite  numbers, 
the  work  can  uhcays  be  contracted. 

Example  :  Multiply  375  by  7,  and  divide  that  product  by  21.  To 
obtain  the  answer,  it  is  sufficient  to  divide  375  by  3,  which  gives  125. 

'i  he  7  divides  the  21,  and  the  foctorS  remains  for  a  divisor.  Here 
it  becomes  necessary  to  lay  down  a  plan  (f  operation. 

Dravy  a  perpendicular  line,  and  place  all  numbers  that  are  to  be 
multiplied  together  under  each  other,  on  the  right  hand  side,  and  all 
numbers  that  are  divisors  under  each  other,  on  the  left  hand  side. 

EXAMPLES. 

1.  Multiply  140  by  36,  and  divide  that  product  by  84.  We  place 
the  numbers  thus ; 

We  may  cast  out  equal  factors  from  each  side  of  the  line  without 
affecting  the  result.  In  this  case,  13  will  divide  84  and  36;  then, 
the  numbers  will  stand  thus: 

140 
3 

But  7  divides  140,  and  gives  20,  which,  multiplied  by  3,  gives  60 
for  the  result. 

2.  Multiply  4783  by  39,  and  divide  that  product  by  13. 

4783 
^'^  3 
Three  times  4783  must  be  the  result. 

3.  Multiply  80  by  9,  that  product  by  21,  and  divide  the  whole  by 
the  product  of  60X6X14. 


^^ 


3  ^0 

6 

2^^ 


^0  4 
9 
#  3 


In  the  above,  divide  60  and  80  by  20,  and  14  and  21  by  7,  and 
those  numbers  will  stand  canceled  as  above,  with  3  and  4,  2  and  3  at 
their  sides. 


48  ARITHMETIC. 


Now,  the  product  3X6X2,  on  the  divisor  side,  is  equal  to  4  limes 
9  on  the  other,  and  the  remaining  3  is  the  result. 

Hoping  now  that  the  pupil  understands  our  forms,  and  compre- 
hends the  true  philosopliical  principles,  we  give  a  few  unwrought  ex- 
amples for  exercises. 

4.  Multiply  84  by  56,  and  divide  the  product  by  14 ;  what  is  the 
result  ? 

5.  What  is  the  result  of  75X21,  divided  by  7  1 

6.  What  is  the  result  of  126X72,  divided  by  48  1 

7.  What  is  the  result  of  5728X49,  divided  by  56? 

8.  What  is  the  result  of  64X  18X48,  divided  by  16X9X13? 

9.  What  is  the  result  of  125X8X2,  divided  by  100X24  ? 

10.  What  is  the  result  of  39X41X360,  divided  by  82X30? 

1 1.  What  is  the  result  of  22AX  13X37^,  divided  by  75X45? 

12.  What  is  the  result  of  71X19X7,  divided  by  38X211 

13.  What  is  the  result  of  221X635,  divided  by  35X5  ? 

APPLICATION. 

1.  A  farmer  sold  28  bushels  of  wheat  at  85  cents  per  bushel,  and 
took  his  pay  in  cotton  cloth  at  14  cents  per  yard ;  how  many  yards 
did  he  receive?  -4ns,  170. 

2.  A  person  bought  12  yards  of  cloth  at  219  cents  per  yard,  and 
paid  in  butter  at  9  cents  per  pound  ;  how  many  pounds  did  it  require  ? 

Ans.  292. 

3.  How  many  yards  of  cloth,  at  $4.66  a  yard,  must  be  given  for  18 
barrels  of  flour  at  §9.32  a  barrel  ?  A/is.  36. 

4.  The  children  in  a  Sunday-school  contributed  5  dollars  to  a  chari- 
table object ;  each  giving  6i  cents ;  how  many  children  were  there  ? 

Ans.  80. 

5.  A  laborer  vvrorked  26  days  at  87^  cents  per  day,  and  look  his 
pay  in  wheat  at  65  cents  per  bushel ;  how  many  bushels  did  he  re- 
ceive ?  Ans.  35, 

6.  A  person  pays  3^  dollars  a  week  for  board  ;  how  many  dollars 
must  he  pay  for  26  weeks?  Ans.  13  times  7:  $91. 

7.  A  merchant  bought  526  barrels' of  flour  at  $4  50  per  barrel,  and 
paid  in  cloth  at  $2.25  per  yard  ;  how  many  yards  did  it  require  ? 

Ans.  1052. 

8.  How  much  land,  at  §2.50  per  acre,  must  be  given  in  exchange 
for  360  acres  at  §3.75  per  acre  ?  An^s.  540. 

9.  What  will  28  pounds  of  sugar  cost,  at  9|  cents  per  pound? 

Ans.  7  times  39  cents :  or,  §2,75. 

10.  An  auctioneer  sold  55  bags  of  cotton,  each  containing  400 
pounds,  receiving  1  mill  commission  on  a  pound  ;  how  many  dollars 
did  his  commission  come  to  ?  Ans-  §22. 

11.  How  many  casks,  each  containing  1  bushel  I  peck,  are  required 
to  hold  145  bushels]  Ans.  116. 

12.  How  much  will  540  yards  of  cloth  cost  at  3  shillings  4  pence 
per  yard,  in  dollars,  at  6  shillings  each  1  Ans.  §300. 


MULTIPLICATION    AND    DIVISION.  49 


13.  If  1  quart  cost  10  pence,  how  many  pounds  will  IJi  hogsheads 
cost]  ybr.  £126. 

14.  How  many  pounds  of  butter  at  9^  cents  per  pound,  will  pay 
for  19  yards  of  muslin  at  II  cents  per  yard  ?  Aus.  22. 

15.  How  many  bushels  of  oats  at  22^  cents  a  bushel,  will  pay  for 
75  pounds  of  sugar  at  5  cents  per  pound  ?  An.".  16-f. 

16.  How  many  bushels  of  wheat  at  75  cents  a  bushel,  will  pay  for 
6  yards  of  cloth  at  3^  dollars  a  yard  ?  Ans.  28. 

17.  How  many  bushels  of  corn  at  45  cents  per  bushel,  will  pur- 
chase 24  yards  of  carpeting  at  65  cents  per  yard  ]  Ans.  34^. 

18.  How  many  tons  of  hay  at  5^  dollars  per  ton,  will  pay  for  11 
acres  of  land  at  19  dollars  per  acre?  Ans.  ;38. 

19.  If  a  man  travel,  on  an  average,  4  miles  an  hour  and  9  hours  a 
day,  how  many  days  will  be  required  to  pass  over  1260  miles  ? 

Ans.  35. 

20.  How  many  bushels  of  barley  at  75  cents  per  bushel,  will  be  re- 
quired to  [);iy  a  debt  of  ^45.75?  Ans.  61. 

21.  How  many  days'  work  at   125  cents  per  day,  will  pay  for  80 
acres  of  land  at  225  cents  per  acre  ?  Ans.  144, 

22.  What  number,  multiplied  by  23,  will  give  the  same  product  as 
75  multipUed  by  691  "  Ans.  225. 

23.  What  number,  multiplied  by  16-^,  will  give  the  same  product 
as  300  multiplied  by  451  ?  Ans.  8200. 


COMPOUND  NUMBERS. 

(AuT.  25.)  Compound  Numbeus  are  such  as  express  quantities 
consisting  of  different  denominations,  but  of  the  same  general  kind, 
such  as  bushels,  pecks,  quarts,  &c. ;  yards,  feet,  inches,  &C.  The 
most  proper  appellation  for  these  quantities  is  Denominnte  Numbers, 
because  they  simply  consist  of  dijftrcnt,  denoniinutions,  and  are  not 
comfiound ;  but  the  name  compound  numbers  has  been  so  long  at- 
tached to  them  that  it  would  be  dilncult  to  cliange  it. 

The  following  Tables  of  the  denominations  of  compound  numbers 
are  to  lie  committed  to  memory  before  entering  upon  reduction. 

ENGLISH  MONEY. 

The  denominations  of  English  Money  are,  guineas,  pounds,  shil- 
lings, pence,  and  farthings. 


4  firthings,  marked  far.,  make  1  penny,  marked  d. 

12  pence 1  shilliiig,  .    .    .   s, 

20  shillings 1  pound,  .    .    .    £. 

21  shillings 1  guinea. 


50  ARITHMETIC. 


MENTAL    EXERCISES. 

In  £2  3s.,  how  many  shillings  ? 
In  2.S-.  2d.,  how  many  pence  ? 
In  35.  2f/.,  how  many  pence  1 
In  £Z  2s.,  how  many  shillings  ? 

It  is  evident,  from  the  inspection  of  the  table,  that  to  change  pounds 
to  shillings  we  must  multiply  the£l  by  20,  and  to  change  shillings  to 
pence  we  must  multiply  the  shillings  by  12,  &c. ;  and  this  changing 
of  a  quantity  from  one  denomination  to  another  is  called  Reduction  ,- 
for  it  is  reducing. 

TROY  WEIGHT. 

Gold,  silver,  jewels,  and  liquors,  are  weighed  by  this  weight.  Its 
denominations  are  pounds,  ounces,  pennyweights,  and  grains. 

TABLE. 

24  grains,  gr.,  .    .    .  make  1  pennyweight,  marked  dwt. 

20  pennyweights  ....     I   ounce, oz. 

12  ounces 1  pound, lb. 

MENTAL    EXERCISES. 

In  2  pounds  1  ounce,  how  many  ounces  ? 

In  2  ounces  3  pennyweights,  how  many  pennyweights? 

In  2  pennyweights  and  2  grains,  how   many  grains? 

In   1  pound,  how  many  pennyweights  ? 

In  1  ounce,  how  many  grains  ? 

APOTHECARIES'  WEIGHT. 

This  weight  is  used  by  apothecaries  and  physicians  in  mixing  their 
medicines.  Its  denominations  are  pounds,  ounces,  drams,  scruples,  and 
grains.  The  pound  and  ounce  are  the  same  as  the  pound  and  ounce 
in  the  Troy  weight;  the  difference  belween  the  two  weights  consists 
in  the  different  divisions  and  subdivisions  of  the  ounce. 

TABLE. 

20  grains,  gr., make   1  scruple,  marked  g 

3  scruples, 1  dram 3 

8  drams, 1  ounce  .    .    .    .    g 

12  ounces, 1  pound  ....  lb 

AVOIRDUPOIS   WEIGHT. 

By  this  weight  are  weighed  all  coarse  articles,  such  as  hay,  grain, 
chandlers'  wares,  and  all  the  metals,  excepting  gold  and  silver.  Its 
denominations  are  tons,  hundreds,  quarters,  pounds,  ounces,  and 
drams. 

'i'he  hundred  weight  is  1 12  pounds,  as  appears  from  the  table;  but 
at  the  present  time  the  merchants  in  our  principal  cities  buy  and  sell 
by  the  100  pounds,  and  20  hundreds  a  ton. 


TABLES.  51 


TABI-E. 

16  drams,  dr., make  1   ounce,  marked  oz. 

1  ()  ounces 1   pound,  .    .    .   .  lb. 

23   pounds, 1   quarter,  .    .    ,     cp: 

4  quarters, 1   Jmndred  weight,  cict. 

20  hundred.  , 1  ton, T. 

MENTAL   EXERCISES. 

In  10  pounds,  how  many  ounces  ? 
\n  2  pounds  8  ounces,  how  many  ounces  ? 
In  3  pounds  2  ounces,  how  many  ounces? 
Ill  10  ounces  10  drams,  how  many  drams  1 
In  100  pounds,  how  many  ounces'? 

LONG   MEASURE. 

Tii>i!sr  measure  is  used  when  length  only  is  considered.  Its  denomi- 
nations are  degrees,  leagues,  miles,  furlongs,  rods,  yards,  teet,  inches, 

and  barley-corns. 

TABLE. 

3  harlev-corns,  bar.,  .    .    .  make  1  inch,  .    .    .  marked  in. 

12  inche's, 1  foot, ff, 

3  feet 1  yard, yd. 

5h  yards,  or  16^  feet, 1  rod,  perch,  or  pole,  rd. 

40  rods, 1  furlong, fur. 

8  furlongs,  or  320  rods,  ....  1  mile, 7ni. 

3  miles, 1  league, L. 

60  geoi^raphical  or  69A  statute  ">     ,     ,  j  o 

°     °    ^    .,  ■'■  5-1   decree, di:s^.ox° 

miles,  5  •••"•& 

orn  1  C  a  great  circle,  or  circumference 

360  degrees, <     °  ..  ,. 

=>       '  C      ol  the  earth. 

[Note.  A  fathom  is  six  feet,  and  is  generally  used  to  measure  the 
depth  of  water. 

A  hand  is  four  inches,  and  is  used  to  measure  the  height  of  horses.] 

MENTAL    EXERCISES. 

How  many  inches  in  one  yard  ? 

How  many  inches  in  2  feet  2  inches  ? 

How  many  feet  in  2  rods? 

How  many  feet  in  2  rods  and  two  feet? 

In  3  leagues  and  I  mile,  how  many  miles  ] 

CLOTH  MEASURE. 

Cloth  measure  is  used  for  measuring  all  kinds  of  cloth.  Its  de- 
nominations are  ells  French,  ells  English,  ells  Flemish,  yards,  quar- 
ters, nails,  and  inches. 


52  ARITHMETIC. 


TABLE. 

2^  inches,  in., make  1  nail,  .    .    .  marked  na. 

4  nails, 1  quarter  of  a  yard,  .  qj: 

4  quarters, 1  yard, yd. 

3  quarters, 1   Ell  Flemish,  .    .  .  E.  Fl. 

5  quarters,  • 1   Ell  English,  .    .  .  E.E. 

6  quarters, 1  Ell  French,  .    .    ^  E.  Fr. 

MEXTAL    EXERCISES. 

In  4^  inches,  how  many  nails  ? 
In  2  quarters  2  nails,  how  many  nails? 
In  3  yards  3  quarters,  how  many  quarters  ? 
In  2  yards  2  quarters,  how  many  quarters? 
In  i  yard  2  nails,  how  many  nails] 

LAND  OR  SQUARE  MEASURE. 

Land  or  square  measure  is  used  in  measuring  land,  or  anything  in 
which  length  and  breadtli  are  both  considered. 

TABLE. 

144  square  inches,  sg.  in.,  .    .    .  make  1  square  foot,  .  sq.  ft. 

9  square  feet I  square  yard,.,  .sv^.j/q'. 

30;^  square  yards, 1  square  pole,  .  F. 

40  square  poles, 1  rood,  .    .    .      A'. 

4  roods, 1  acre,  ,    .    .  .  A. 

640  acres, 1  square  mile,  .  M. 

The  surveyor's  or  Gunter's  chain  is  generally  used  in  surveying 
land.  It  is  four  poles,  or  66  feet,  in  length,  and  is  divided  into  100 
links. 

TABLE. 

7jW  inches make   1   link,  ....  marked  /. 

4   rods,  or  66  feet,  ....     1   chain, c. 

80  chains 1   mile, mi, 

I   squ  ire  chain, 16  square  poles,  .    .    .  .   P. 

10  square  chains, 1   acre, A. 

Land  is  generally  estimated  in  square  miles,  acres,  roods,  and  square 
poles  or  perches.* 

MtXTAL  EXERCISES. 

In  2  feet  square,  how  many  square  feet  ?  Ans.  4.     i , r 

A    surface  3  feet   in    length   and   2  feet  in  width— how  I j 

many  square  feet  ?  '  I  i 
A  surface   6  feet  in  length  and  3  in  width,  how  many  i ' I 

square  feet  ? 

♦More  recently  in  mileB,  acres,  and  decimals  of  an  acre. 


TARLK 


53 


A    towiishii)  6  miles  long  and  4   i 
miles  ciiX's  it  coiitain  '.' 

How  many  square  miles  in  10  miles  square  ] 


bruat!,  how  many  square 


SOLID  OR  CUBIC  MEASURE. 

8olid  or  cubic  measure  is  used  in  measuring  such  things  as  have 
three  (limensions,  of  length,  breadth,  and  thickness.  Its  denomina- 
tions are  tons,  cords,  yards,  feet,  and  inches. 


1728  solid  inches,  s.  in.,  .    . 

'27  solid  feet,   • 

40  feet  of  round,  or         ') 

50  feet  of  hewn  timber,  5 


make  1  solid  fool,  marked  s.  fl, 

s-  yd. 


1   solid  yard,  . 
I   ton,  . 


Ton. 


12S  solid  feet=8X4X't,  that  is,  ~) 
a    pile    8    feet   in  length,  4  feel  in  >    1   cord  of  wood,  .    .  C. 
width,  and  4  feet  in  height,  ^ 

[NoTK.  A  cord  foot  is  1  foot  in  length  of  the  pile  which  makes  a 
cord.     It  contains  1 6  solid  feet.] 

MENTAL    EXERCISES. 

A  block  1  foot  long,  1  wide,  and  1  high,  is  a  cubic  foot.  A  block 
1  foot  square  at  the  end  and  M  feet  long,  how  many  cubic  feet? 

A  block  containing  4  s([uare  inches  at  the  end  and  10  inches  bng, 
how  many  cubic  inches? 

WINE  MEASURE. 

Wine  iriejisure  is  used  in  measuring  ail  liquors,  excepting  beer  and 
ale.  Its  denominations  are  tuns,  pipes,  hogsheads,  barrels,  gallons, 
quarts,  pints,  and  gills. 

TABM-. 


4  trills,  o-,-'., 

2  pint.s,\    . 

4  quarts,  . 

3  4  gallons,  . 

63  gallons,  . 


make  1  pint, 
.    .  .    1  quar 


;  gallon,  . 
1  barrel,  . 
I  hogshead, 


2  hogsheads, 1  pq;e. 

2  pipes,  or  4  hc)gsheads,  .    .    . 


tun, 


marked  pt. 
.    .    .      gt 
.    .    .    gal. 
....  oar. 

.  Md. 

.  pi. 
tun. 


A  wine  gallon  contains  23 1  cubic  inches. 

MENTAL   EXERCISES. 

In  2  pints,  how  many  gills  ? 

In  3  pints  1  gill,  how  many  gills? 

In  3  gallons,  how  many  quarts? 

In  3  gallons,  2  quarts,  I  pint,  how  many  pints? 

In  1  gallon  1  pint,  how  many  pints  ? 


54  ARITHMETIC. 


ALE  OR  BEER  MEASURE. 

Ale  or  beer  measure  is  used  in  measuring  ale,  heer,  and  milk.  Its 
denominations  are  hogsheads,  barrels,  gallons,  quarts,  and  pints. 

TABLE, 

2  pints,  pt., make  1  quart,  .    .  marked  qt. 

4  quarts, 1  gallon, gal 

36  gallons, i  barrel, bar. 

54  gallons, 1  hogshead,  .    .    .    hhd. 

A  gallon,  beer  measure,  contains  282  cubic  inches. 
DRY  MEASURE. 

Dry  measure  is  used  in  measuring  all  dry  articles,  such  as  grain, 
fruits,  roots,  salt,  coal,  &c.  Its  denominations  are  caldrons,  bushels, 
pecks,  quarts,  and  pints.] 

TABLE. 

2  pints,  pt make  I  quart,  .  .  .  marked  qt. 

8  quarts, 1  peck, pk. 

4  pecks, 1  bushel, hu. 

36  bushels,  ....     • 1  caldron, cul. 

[Note.  A  gallon;  dry  measure,  contains  2684  cubic  inches.  A 
Winchester  bushel  is  18^  inches  in  diameter,  8  inches  deep,  and  con- 
tains 21501  cubic  inches. 

5 

A  bushel  of  coal,  lime,  &c.,  is  2688  cubic  inches. 
TIME. 

TABLE. 

60  seconds,  sec, make  1  minute,  .    .  marked  mm. 

60  minutes, 1  hour, h. 

24  hours, 1  day, d. 

7  days 1  week, w. 

12  months,  (or  365  days,)  ...  1  year, y, 

[Note.  The  true  year,  according  to  the  latest  and  most  accurate 
observations,  consists  of  365  days,  5  hours.  48  minutes,  and  58  seconds  ; 
this  amounts  to  nearly  365^  days.  The  common  year  is  reckoned 
365  days,  and  every  fourth  or  leap  year  one  day  more. 

Those  years  which  are  divisible  by  4,  without  remainders,  are  leap 
years.  By  estimating  5  hours,  48  minutes,  58  second:=,  as  6  hours, 
accumulated  an  error,  which  amounted  to  more  than  1 1  days  in  1752 ; 
and  the  II  days  were  dropped  to  correct  the  style  :  hence,  before  that 
period  is  called  Old  Style,  subsequently  New  Style. 

A  further  correction  of  one  day  took  place  in'  the  year  1800  ;  still 
a  further  correction  will  take  place  in  the  year  1900,  after  which  there 
will  be  no  correction  for  200  years.] 


REDUCTION. 


55 


MENTAL   EXERCISES. 

In  2  hours  10  minutes,  how  many  minutes? 

In  3  minutes  10  seconds,  how  many  seconds? 

In  10  hours,  how  manyminutesl 

In  three  hours,  how  many  minutes?  how  many  seconds? 

In  2  hour.--,  how  many  minutes? 

In  13  minutes  3  seconds,  how  many  seconds  1 

CIRCULAR  MEASURE,  OR  MOTION. 
Circular  measure  is  used  in  calculating  latitude  and  longitude,  and 
also  in  measuring  the  motions  of  the  heavenly  bodies.      Every  circle 
is  supposed  to  be  divided  into  360  equal  parts,  called  degrees. 

TABLE. 

60  seconds  (") make  1  minute,  ,    .  marked 

60  minutes, 1  degree, " 

30  degrees 1  sign, s. 

12signs,  or3()0° 1  circle c. 

The  signs  are  becoming  obsolete.     'l"he  best  astronomers   of  the 
present  day  do  not  use  them. 

MENTAL    EXERCISES. 

In  1  sign  10  degrees,  how  many  degrees? 

In  2  degrees,  how  many  minutes? 

In  3  degrees  3  minutes,  how  many  minutes  ? 

In  2  signs  2  degrees,  how  many  degrees  ? 

How  many  minutes  in  a  quadrant,  or  90  degrees? 

PARTICULARS. 

12  units  make a  dozen. 

12  dozen, a  gross. 

144  dozen, a  great  gross. 

20  units, a  score. 

24  sheets  of  paper, a  quire. 

20  quires, a  ream. 

BOOKS. 

A  sheet  folded  in  two  leaves  is  called  a  folio. 

folded  in  four  leaves  ....   a  quarto,  or  4to. 
'<       folded  in  eight  leaves  ...    an  octavo,  or  8vo. 
"      folded  in  twelve  leaves  ...ad  lodecimo,  or   12mo. 
•*      folded  in  eighteen  leaves  .    .  an  18mo. 


REDUCTION". 


(Art.  26.)     Redcctiov  is  reducing  or  changino;  one  dcnomina. 
tion  of  any  quantity  to  another  denomination,  without  changing   its 


56  ARITHMETIC. 


real  value;  and  it  invoivos  no  other  principles  than  those  called  out 
ill  the  mental  exercises  under  the  tal-.les. 

From  these  exercises  we  );erceive  that  quantities  may  be  reduced 
from  a  higher  to  a  lower  denomination,  l»y  mnltiphcation. 

Thus,  2  yards  are  6  feet:  these,  reduced  to  inches,  give  72. 

Conversely,  lower  denominations  may  be  brought  to  higher,  by 
division,  as  pounds  to  qu  irlcrs,  hundreds  weight,  or  tons.  In  all  these 
cases,  the  quantity  remains  unchanged,  although  it  is  expressed  in  a 
diflrrcnt  denomination. 

When  higher  denominations  are  to  be  reduced  to  lower,  the  expla- 
nation gives  the  following 

r?iTT,K.  Mnlliply  iJie  successive  denominafions,  comincndns;  iciih 
Ihe  hig.'ie-f  given,  by  the  number  iliai  ivill  make  if  one  of  the  next 
hncer,  adding,  in  their  proper  placea,  ihe  several  inferior  denomina- 
tions expressed  in  the  given  number. 

When  lower  denominations  are  to  be  reduced  to  higher,  the  expla- 
nation gives  the  following 


Rule.  Divide  the  given  denomination  by  the  number  that  will 
make  one  of  the  next  higher,-  and  the  quotient  thence  arising  by  as 
manyas7nake  one  of  ike  denomination  next  above  that,  and  soon 
to  ihe  required  one. 

The  farmer  of  these  rules  is  reduction  descending,  the  latter 
ascending.     Being  reverse  operations,  they  pnjve  each  other. 

As  the  principles  of  the  operation  are  the  same  under  each  table  of 
weight,  measure,  or  motion,  we  shall  give  the  examples  promiscuously, 
to  exercise  quickness  of  thought. 


1,  Reduce  17  lbs.  W  oz.  15  diets.,  to  pennyweights,* 

nibs. 
:  1^  cz. 


:o4 

1 1  oz.  added. 


In   this  example,  we  first  multiply  by  | 
the  number  of  ounces  in  a  pound,  and  215 

then  add  the  ounces,  &c,,  «Scc. 


20  dwi. 

4300 

15  dwts.  added. 

4315  dwts. 


*  A  practical  man  would  at  once  perceive  that  this  is  5  pennyweights  less 
than  18  pounds  :  therefore,  18 X  12X20— 5=the  result. 


REDUCTION 


57 


2.  In  2.')  l/js.  9  oz.  0  f/a'/i^  10  gr.,  how  many  grains? 

Ans.   148330. 

3.  RediJCfi  7461  fi  farthings  to  pounds. 

4) 71016 


12)18654  pence. 


20)1554  shillings  and  6  pence. 

77  and  14  shillings  over. 

Am.  £77,   145.  6d=746 16  farthings. 

4.   In  3278  nails,  how  many  yards? 

4)3278 


We  first  divide  by  4,  which  brings 
the  number  to  quarters,  and  then  again 
by  4,  which  brings  it  to  yards. 


4)819  .  2  n«. 
204  .  3  qrs. 


A)ts.  204  yds.  3  qrs.  2  7ia. 
5.   In  £1234  155.  7c?.,  how  many  farthings? 
£      s.   d. 
'234  15  7 
20 


24695  shillings,  15  being  added  in  mentally. 
12 


296317  pence,  7  being  added  in  mentally. 


.4/15.   1185388  farthings. 


6.  Reduce  3060288  cubic  inches  to  tons  of  round  timber. 

1728)3060288(1771 


We  first  divide  by  1728,  the  num- 
ber of  solid  inches  in  a  solid  foot,  and 
next  by  40,  the  number  of  solid  feet 
in  a  ton 


410)177(1 

44     1 1 


1728 

13322 
12096 

12268 
12096 

1728 
1728 


Ans.  44  tons  11  feet. 


58 


ARITHMETIC. 


7.  Reduce  43  gallons  3  quarts  1  pint,  to  pints. 
43 
4 


175 


Ans.  351  pints. 

(AiiT.  27.)     8.  Reduce  35  tons  to  drams. 

When  there  are  no  inferior  denominations,  as  quarters,  pounds,  &c., 
to  add  in  as  the  operation  advances,  we  may  arrange  all  our  multi- 
pliers before  the  eye,  and  take  the  product  of  several  of  them  men- 
tally, to  shorten  the  operation. 

In  the  present  example  we  multiply  35  tons 

by  20 
by  4 
by  28 
by  16 
by   16 

Then,  78400 

by  256       gives  the  answer. 


35X20=700 

4X28=112 

16X16=256 


470400 
3920 
1568 

20070400 

9.  Reduce  23  bushels  3  pecks  5  quarts  1  pint,  to  pints. 

N.  B.  This  is  5  pints  less  than  24  bushels  ;  therefore,  if  we  reduce 
24  bushels  to  pints,  deducting  5,  we  have  the  answer. 

Multiply  24 
by    ■'    4 

by  8 

by  2 


10.  Reduce  H  acres  3  roods  15  rods,  to  rods.  Ans.  1895. 

1 1.  Reduce  24  square  rods  to  square  feet.  Ans.  6534. 

12.  Reduce  10  pipes  1  hogshead  and  2  quarts  of  wine,  to  pints. 

Ans.  10588. 

13.  Reduce  3  hogsheads  to  gills.  Ans.  6048. 


That  is  .    .    . 

....  64 

by 

.    .    .  .  24 

128 

256 

1531  Ans. 

REDUCTION.  59 


14.  In  1746880  ounces,  how  many  tons, 

Ans.  48  tons,  and  14  cwt.  3  qrs.  8  lbs.  over. 

15.  In  17645  grains,  apothecaries'  weif^ht,  iiow  many  pounds  ? 

Ans.  3  Ib.s.,  and  (i  dr.  0  scru.  5  grains  over. 

16.  In  168474  feet,  how  many  miles'! 

An'\  31  n)iles,  and  7  fur.  58  yds.  over. 

17.  In  2419200  seconds,  how  many  daysl  '"  Ajis.  28. 

18.  In  547325  fartliings,  how  many  pence,  shillings,  and  pounds? 

Ans.  £579  2s.  Gii.  1  qr. 

19.  In  9173  nails,  how  massy  yards?     Ans.  573  yds.  1  qr.  1  n. 

20.  How  many  barleycorns  will  reach  rovnid  the  earth,  supposing 
it,  according  to  the  best  calculations,  to  be  24877  miles  ? 

A}is:  4728G20160. 

21.  How  many  seconds  are  in  a  solar  year,  or  365  days  5  hours  48 
minutes  58  seconds?  Ans.  .j1556938. 

22.  In  a  lunar  month,  or  29  days  12  hours  44  minutes  3  seconds, 
how  many  seconds]  A7is.  2551443. 

23.  In  an  ingot  of  silver  weighing  6  pounds  3  ounces  10  gnins, 
how  many  grains?  Ans.  36010. 

24.  In  5  hundreds  2  quarters  16  pounds,  how  many  pounds  ? 

Ans.  G32. 

25.  In  4  miles  6  furlongs  22  rods,  how  many  rods  ?     Ans.  1542. 

26.  In  11456  geographic  miles,  how  many  degrees  ? 

Ans.   190  deg.  56  min. 

27.  In  164736  inches,  how  many  miles'? 

Ans.  2  miles  4  fur.  176  yds. 

28.  In  14764  nails  of  cloth,  how  many  yards  ? 

Ans.  922  yds.  3  qr. 

29.  In  2746  quarters,  how  many  ells  English  ? 

A71S.   549  ells   1   qr. 

30.  Reduce  2  tuns  to  gills.  Ans.  IG\28. 

31.  Reduce  32  gallons  3  quarts  to  pints.  Ans.  262. 

32.  Reduce  2  hogsheads  27  gallons  3  quarts,  to  quarts. 

Ans.  615. 

33.  Reduce  3  tuns  I  hogshead  15  gallons  1  quart,  to  pints. 

Ans.  CG74. 

34.  Reduce  1484  yiints  to  gallons.  Ans.  185  gal.  2  qts. 

35.  Reduce  167000  gills  to  barrels  of  32  gallons  each. 

Ans.  163  barrels  2  gals.  3  qfs. 

36.  Reduce  6272640  square  inches  to  acres.  Ans.  1. 

37.  A  grocer  purchased  16  hundred  weight  3  quarters  20  pounds 
of  sugar  ;  how  many  pounds  are  there  in  the  whole?      Ans.  1896. 

33.  A  tobacconist  bought  116  hundred  weight  3  quarters  of  tobac- 
co; how  many  pounds  were  in  the  whole  ?  Ans.  13076, 

39.  A  silversmith  has  7  pounds  10  ounces  of  silver;  how  many 
grains  are  in  the  whole  ?  Ans.  45120. 

40.  An  apothecary  has  several  sorts  of  drugs,  weighing  toj^ether  47 
pounds  6  ounces  4  drams;  how  many  scruples  and  grains  arc  in  the 
whole?  Ans.   13692  scru.,  273810  grs. 


60  ARITHMETIC. 


41.  A  merchant  wishes  to  ship  300  bushels  of  flaxseeJ  in  casks  con- 
taining 7  bushels  2  pecks  each ;  what  number  of  casks  are  required  1 

Atis.  40. 

42.  A  certain  barn  is  64  by  36  feet,  and  20  feet  high  ;  another  bam 
is  32  by  24  feet,  and  16  feet  high:  how  much  larger  is  the  former 
than  the  latter  ?     (8ee  solid  measure.)  Ans.  3^  times  larger. 

43.  How  many  times  will  a  wheel  16  feet  6  inches  in  circumfer- 
ence, turn  round  in  running  42  miles  ?  A/is.  13440  times. 

44.  There  are  two  places  on  the  equator,  one  in  longitude  40  deg. 
east  of  the  meridian  of  Greenwich,  tlie  other  30  deg.  west;  what  is 
the  number  of  geographical  miles  between  them  ?  An.^.  4200. 

45.  A  ship,  during  3  days  of  storm,  had  changed  her  longitude  273 
geographical  miles  ;  how  many  degrees  and  minutes  of  a  degree  did 
she  change  ]  Ans.  4  deg.  33  min. 

46.  How  many  quires  and  reams  of  paper  are  in  500520  sheets  ? 

Ans.  20855  quires,  1042  reams  15  quires. 

47.  A  printer  calls  for  4  reams  10  quires  and  10  sheets  of  paper  to 
print  a  book  ;  how  many  sheets  does  he  call  for  ?  Ans.  2170. 

48.  How  many  solid  inches  are  in  a  solid  7  inches  long,  7  inches 
wide,  and  7  inches  high  1  Ans.  343. 

49.  How  miny  solid  inches  are  in  a  solid  2  feet  1  inches  long,  6 
inches  wide,  and  6  inches  high  1  Ans.  27X30. 

50.  How  many  solid  inches  are  in  a  solid  3  .feet  2  mches  long  by  2 
feet  2  inches  wide  and  1  foot  8  inches  high? 

A?is.  38X26X20=19760. 

51.  How  !n;»ny  yards  of  carpeting,  1  yard  wide,  will  be  required  to 
carpet  a  room  18  feet  wide  and  20  feet  long?  Ans.  40. 


-^COMPOUND  ADDITION. 

(Art.  28.)  Compound  Addition  is  simple  addition  and  reduc- 
tion combined  ;  but  we  shall  be  more  clearly  comprehended  through 
the  mi'dium  of  an  example. 

What  is  the  sum  of  £19  15s.  8d.,  £27  18s.  6d.,  £43  9s.  4d.,  and 
£7  6s.  6;]. 

As  unlike  things  cannot  be  added  together,  (.\rt.  5.)  we  must  write 
pounds  under  pounds,  shillings  under  shillings,  and  pence  under  pence, 
thus: 

£  s.  d. 

19  15  8 

27  18  6 

7  6  6 

55       0     8 


UKDUCTION.  01 


Tn'ow,  each  column,  by  itself,  is  a  sum  in  t-implc  !i<lcUtion  ;  and,  as 
luTO  are  three  coliiiniis,  we  have  tliree  sums  in  adililion.  But  when 
we  make  the  reduced  sum  of  one  column  run  inlo  the  next  column 
we  unite  or  caitipuund  Hit  additions. 

We  must  commence  with  tlie  lowest  column,  (the  pence  column)  ; 
its  sufn  is  20  pence,  which,  reduced  to  shillin.^^s.  give  1  .shilling,  and  H 
pence  over.  'i'he  8  pence  must,  of  course,  be  [)ut  down  in  the  penc<> 
roluuu^.  'J'he  I  shilling,  added  in  with  the  other  shillings,  all  make 
40  shillings;  which,  reduced,  make  2  pounds  and  0  shillings  over. 
The  2  pounds,  added  into  the  column  of  pounds,  make  the  whole 
number  of  pounds  55;  and  the  several  sums  of  money  make  the  total 
sum  of  £55  Os.  8d. 

If  the  pupil  retains  the  principle,  that  compound  addition  is  simple 
addition  and  reduction  combined,  no  rule  for  addition  will  be  required  ; 
but  for  the  sake  of  conciseness,  we  give  the  following 

Rule.  1.  Flace  the  numbers  so,  that  those  of  the  same  denomi- 
nation will  stand  directly  under  each  other,  and  in  distinct  and  sepa- 
rate columns. 

2.  Add  up  the  Jis;iires  in  the  lowest  denomination,  and  find,  by 
reduction,  how  many  units  or  ones  of  the  next  higher  denomination 
are  contained  in  their  sum, 

3.  Set  down  the  remainder  below  its  proper  column,  and  carry 
those  units  or  ones  to  the  next  denomination,  which  add  up  in  the 
same  manner  as  before. 

4.  Proceed  thus  through  all  the  denominations,  to  the  highest, 
whose  sum,  together  with  the  several  remainders,  luill  give  the 
answer  sought. 

N.  B.  The  method  of  proof,  rather  a  test,  is  to  vary  the  order  of 
addition  :  that  is.  break  the  sum  into  several  sums,  and  then  take  the 
partial  sums  for  the  sum  total ;  and  if  two  methods  give  the  same  sum 
total,  that  sum  may  be  regarded  as  correct. 


ENGLISH  MONEY. 


£ 

s. 

d. 

48 

13 

8 

51 

6 

4 

67 

11 

3 

76 

18 

10 

244 

10 

1 

EX  A 

MPLE 

s. 

£ 

.<?. 

d. 

876 

12 

8 

542 

11 

9 

213 

6 

4 

457 

16 

7 

£  s.  d. 

124  15  5 

314  12  11 

145  13  2 

278  17  8 


62 

ARITHMETIC. 

1 

LON 
L.    mi.  fur.  rd. 
316     2     7     29 
127     1     2     20 
187     0      1      15 
11      1      1        1 

G  MEASURE. 

TH  MEASUR] 
7/d.  qr.  na. 
*14     3     2 
25     4     1 
16     1     0 
25     2     2 

yd.  ft.    iti.  bar. 
90     2     11     2 
55     1       9     1 
27     0       7     0 
50     2       2     2 

642     2     4 

25 

E.  Fl.  qr.  na. 

26     2     4 
165     2      1 
172      1     3 

57     2     2 

CLO 

OR 

q.  in. 
104 

27 

2 

128 

1 

\SU 
A. 
700 
375 
450 
30 

^.  En.  qr.  na. 

28     5     1 

120     3     1 

219     1     4 

115     3     1 

423     0     2 

SQU 
.   CI 
'E  I\ 

ARE   ME. 

JBIC   ME-^ 
C.     s.ff. 
116     127 
317       12 
418     119 
737     104 

LAND 

sq.  yd.  sq.  ft. . 

197       4 

122       3 

5       8 

237       7 

RE. 
R.  P. 

2  37 

3  25 
1     31 
0     25 

563       5 

117 

SOLII 
5.  yc?.  S.ff.  s.  in. 
l'65     25     1129 
237     26        132 
350       1      1064 
222     19          17 

)    OE 

pt. 
2 
3 
1 
3 
2 

^SU 

pi.  b 
2 

1 
2 
0 

1 

RE, 

C.  cord  ft. 
287     9 
126     7 
316     6 
419     5 

976     18       614 

lEASURE 
tun. 
34 
25 
24 
35 
77 

hhd.  gal.  qt. 

27     65     3 
112     60     2 

50     29     0 
421       0     2 

14     39      1 

hd.  gul.  qt. 

1  27     3 

2  25     2 
1     27     1 

1  62     3 

2  21     2 

627       7     1 

1 

COMPOUND    ADDITION. 

03 

hu.  pk. 

qt.  pi. 

bu.  pk.  qt. 

bu. 

pk.  qt.  pf. 

2:}     3 

7      1 

4     3     7 

3     7      1 

3i     2 

6     1 

5     2     6 

2     6      1 

42     3 

5     1 

6     1     5 

0     5     0 

51      1 

4     1 

7     0     4 

1     4     1 

23     2 

3     1 

8     3     3 

3     3     1 

14     1 

2     1 

4     1     2 

0     2     1 

11     3 

4     I 

3     2     4 

2      1      1 

202     3 

2      1 

40     3     7 
TIME. 

3 

270 

y.    m. 
6     7 

d. 

d. 

h.     m.    sec. 

V' 

m. 

d.     h. 

16 

127 

21     46     55 

0 

10 

21      18 

2     1 

24 

24 

16     30     32 

0 

9 

17       9 

9     9 

21 

17 

12     18     00 

0 

11 

23     23 

4     3 

22 

41 

15     19     12 

0 

2 

9     10 

6     7 

25 

14 

9       9       9 

0 

1 

11      11 

d.   h 

.    m. 

sec. 

d.    h.     m. 

sec. 

h. 

m,  sec. 

6     4     20 

13 

0     7     13 

20 

0 

5     12 

4     2     30 

17 

1     2       9 

12 

0 

6       2 

3     7       8 

20 

2     3       4 

6 

0 

17     45 

7     2     42 

8 

3     4       5 

7 

0 

18      12 



CIRCULAR  MEASURE,  OR  MOTION. 


114 

46 

25 

221 

34 

25 

216 

27 

33 

227 

54 

56 

332 

16 

24 

112 

24 

43 

417 

27 

25 

217 

42 

37 

109 

17 

26 

111 

12 

14 

135 

16 

21 

29 

42 

24 

321 

42 

24 

115 

13 

12 

220 

16 

15 

19 

29 

30 

111 

11 

17 

78 

21 

32 

N.  n.  In  practical  astronomy,  when  the  object  is  to  find  the  posi- 
tion of  a  body  in  a  circle,  we  reject  every  360  degrees  in  addition. 

APPLICATIOX. 

1.  A  merchant  bought  4  barrels  of  sugar;  the  first  weighed  2  hun- 
dred weight  3  quarters  20  pounds,  the  second  1  hundred  weight  2 
quarters  and  15  pounds,  the  third  2  hundred  weight  1  quarter  27 


64  ARITHMETIC. 


pounds, and  the  fourth  2  hundred  weight  3  quarters  and  10  pounds; 
what  was  the  entire  weight  of  the  wholr? 

Alls.   9  cwt.  3  qrs.    IG    lbs. 

2.  A  merchant  shipped  3  boxes  of  freight  to  New  Orleans;  the 
first  weighed  7  hundred  weight  56  pounds,  the  second  4  hundred 
weight  80  pounds,  and  the  third  5  hundred  weight  and  20  pounds ; 
what  was  the  weight  of  the  whole  ] 

This  is  done  Uke  simple  addition ;  modern  reform  is  happily  doing 
away  with  denominate  numbers  in  many  cases. 

3.  There  are  4  fields ;  the  first  contains  12  acres  2  roods  and  3S 
perches,  the  second  4  acres  1  rood  26  perches,  the  third  85  acres  0 
roods  19  perches,  and  the  fourth  57  acres  1  rood  2  perches  ;  how 
many  acres  in  the  four  fields?  Ans.    159a.  2r.  5p. 

4.  In  4  piles  of  wood,  the  first  containing  32  feot  149  inches,  the 
second  121  feet  1436  inches,  the  third  97  feet  498  inches,  the  fourth 
1 15  feet  1356  inches ;  how  much  did  the  whole  amount  to? 

Ans.  2  C.  110  ft.  1711  in. 

5.  In  6  boat-loads  of  wood,  the  first  containing  22  cords  114  feet 
9S7  inches,  the  second  18  cords  121  feet  1436  inches,  the  third  21 
cords  109  feet  1629  inches,  the  fourth  15  cords  82  feet  1321  inches, 
the  fifth  16  cords  98  feet  1111  inches,  the  sixth  2  1  cords  89  feet  987 
inches  ;  how  much  did  they  all  contain  ? 

Ans.  120  C.  105  ft.  559  in. 

6.  A  tailor  requires  1  yard  3  quarters  3  nails  of  cloth  for  a  father's 
coat,  and  1  yard  1  quarter  and  2  nails  for  each  of  2  sons ;  how  much 
in  all?  Ans.  4  yds.  2  qrs.  3  na. 

7.  How  much  land  is  in  a  farm  which  consists  of  50  acres  2  roods 
33  rods  of  wood-land,  25  acres  14  rods  meadow-land.  30  acres  tillage, 
20  acres  rough  pasture  land,  12  acres  I  rood  covered  with  water,  and 
10  acres  25  rods  swamp?  Ans.  148  a.  32  r. 

8.  A  farmer  sowed  3  fields  of  wheat ;  the  first  yielded  45  bushels 
3  pecks,  the  second  36  bushels  1  peck  7  quarts,  and  the  third  30 
bushels  2  pecks  1  quart;  how  much  wheat  did  ho  raise? 

Ans.  112  bu.  3pks. 


COMPOUND    SUBTRACTION. 

(Akt.  29.)  SuBTRACTiov  of  compound  or  denominate  numbers  is 
essentially  the  sarne  as  in  simple  numbers,  except  the  care  that  must 
be  taken  to  borrow  or  reduce  from  the  tables,  in  place  of  borrowing 
or  reducing  from  10.  For  instance,  we  cannot  take  9  pence  from  7 
penre:  we  must  first  add,  or  conceive  to  be  added.  12  pence  more,  (1 
shilling.)  to  7  pence,  borrowing  or  taking  a  shilling  from  the  next  supe- 
rior column,  &c.  From  such  necessary  operations,  we  deduce  the 
folio  win": 


COMPOUND    SUBTRACTION. 


65 


Kci.K.  Place  the  numbers  as  hi  cnrnpotwd  addition.  Begin 
iv:f/i  the  lowest  denominat/oJi  in  the  auhtrahcnd,  and  take  it  from 
the  nutnljer  above  it  ;  but,  if  that  above  be  the  least,  add  as  many  to 
it  as  make  one  of  the  next  higher  denomination  ,-  then  subtract,  and 
carry  one  to  the  next  denomination  in  the  subtrahend. 

The  method  of  proof  is  the  same  as  in  simple  subtraction. 

ENGLISH  MONEY. 


£ 
Borrowed  349 
Paid  .    .     195 


EXAMPLES. 

s.     d.  qr. 
15     6      1 


Lent 


£ 
791 


s.    d.  qr. 
9     8     1 


11     8     1 


Received  197     16     4     2 


Remain;^,    154       3  10     0 


Due  to  me 


AVOIRDUPOIS   WEIGHT. 


T.    cwt.  qr. 
From  45      11      3 
Take    15     10     2 

Rem.    33       I      I 


T.  cwt.  qr.  lbs. 
52  12  3  15 
24     10     0     24 


qr.  lbs.  oz.  dr. 
3     4     0     4 
0     7     9     8 


CIRCULAR  MEASURE,  OR  MOTION. 


From  79°     21' 
Take  41°     41' 


31" 

52'' 


From  147«>     19'     42" 
Take    121^     37'     50" 


Rem.   37°     39'     39" 


A  person  residing  in  latitude  27"  32'  45"  north,  wishes  to  visit  a 
place  52°  24'  18"  north  ;  hou'  many  degrees,  minutes,  and  seconds 
northward  must  he  travel?  Ans.  24°  51'  33". 


y.  m 

6  9 

1  6 

5  3 


TIME. 

w.  d.  h.  m.  sec. 
3  I  2;?  40  20 
2     6     12     57     36 


y.  m.  d.  h. 
9  8  24  47 
6     4     19     39 


From  900  years  take  1 1 1  years  and  6  months. 

Ans.  788  yrs.  6  m. 
If  T  take  1  year  1  month  from  6  years,  what  space  of  time  will  still 
remain  ?  Ans.  4  yrs.  1 1  m. 

_ 


6G  ARITHMETIC. 


CLOTH  MEASURE. 

yd.  qr.  na.                E.  Fl.  qr.  na.  E.  En.  qr.  na. 

35     1     2                     467     3     1  765     1     3 

19     1      3                      291     3     2  149     2     1 


N.  B.  We  deem  it  unnecessary  to  give  formal  examples  corres- 
ponding to  all  the  tables  of  weights  and  measures,  as  the  principles 
of  one  apply  to  all. 

ArPLICATIOX. 

1.  From  3  pounds  8  shillings  and  3  pence,  take  I  pound  7  shillings 
and  9  pence.  Ans.  £2  Os.  6d. 

2.  From  4  pounds  4  pence,  take  19  shillings  5  pence. 

yl;js.  £3  0s.  Ud. 

3.  From  7  pounds,  take  20  grains  Troy. 

Alls.  6  ll)s.  II  oz.  19  dwts.  4  grs. 

4.  From  9  tons  9  hundred  weight,  lake  2  tons  3  hundred  weight  3 
quarters  and  3  pounds.  Ans.  7  T.  5  cwt.  0  qrs.  27  lbs. 

5.  From  3  years  4  months  20  days  16  hours,  take  1  year  5  months 
21  days  14  hours.  '    Ans.  1  yr.  10  mo.  28  d.  2  h. 

6.  What  interval  of  time  elapsed  between  July  12th,  1827,  and 
November  17tli,  1831?  Ans.  4  yrs.  4  mo.  5  d. 

7.  What  is  the  interval  of  time  between  December  23d,  1798,  and 
February  21st.  1811  ?  Ans.  12  yrs.  2  mo.  28  d. 

8.  'J'he  apparent  revolution  of  the  sun  is  365  days  5  hours  48  min- 
utes 58  seconds,  that  of  the  moon  is  27  days  7  hours  43  minutes  3 
seconds  ;  what  is  the  difference  ?        Ans.  337  d.  2L  h.  5  m.  .'J4  sec. 

9.  What  is  the  difference  of  longitude  between  two  places,  one  75° 
21'  30"  west,  and  another  71°  20'  3.5"  west?       Ans.  3°  51'  .55". 

10.  What  is  the  difference  of  longitude  between  two  places,  one 
situated  3°  4'  '20"  east  of  an  assumed  meridian,  and  the  other  1°  20' 
2"  west  of  the  same  1  Ans.  10°  24'  22". 

11.  What  is  the  difference  of  latitude  between  two  places,  one 
latitude  5"^  south,  the  other  5°  north  ?  Ans.  10°. 

12.  What  is  the  difference  of  cold  between  3°  below  zero,  and  10° 
above?  Ans.  13°. 

N.  B.  The  last  three  problems  have  the  algebraic  peculiarity  of 
subtraction,  the  line  of  demarcation  being  between  them. 


COMPOUND  MULTIPLICATION. 

(^Art.  30.)  CoMPOuxD  MuLTiPLTCATToy  IS  thc  same  in  relation 
to  simple  multiplication,  as  compound  addition  is  to  simple  addition : 
that  is,  each  denomination   of    the  multiplicand  is  a  sum  in  simple 


COMPOUND    MULTIPLICATION.  67 


multiplication,  and  its  product  may  require  reduction.     Il  is,  Ihtre- 
fure,  simple  mull ipUcation  and  reduction  combined. 

From  the  above  principle,  the  pupil  will  recognise  the  rationale  of 
the  following 

Rule.  1.  Set  down  the  coinpaund  sum  to  be  miilliplied,  and  un- 
der its  lowest  denomination  at  tlie  right  hand  set  the  multiplier. 

2.  Multiply  the  number  in  the  lowest  denomination  by  the  mul- 
tiplier, and  find  how  many  units  of  the  next  higher  denomination 
are  contained  in  the  product,  setting  down  what  remains. 

3.  In  like  manner  multiply  the  number  in  the  next  denomina- 
tion, and  to  the  product  carry  or  add  the  units,  before  found,  and 
find  how  many  xmits  of  the  next  higher  denomination  are  in  this 
amount,  which  carry  in  like  manner  to  the  next  product,  setting 
doiun  the  overplus. 

4.  Proceed  thus  to  the  highest  denominatio7i  proposed :  so  shall 
the  last  product,  ivilh  the  srceral  remainders,  taken  as  one  common 
number,  be  the  whole  anh/unt  required. 

The  method  of  procf  i^  the  same  as  in  simple  multiplication. 

EXAMPLES. 

bu.  pk.  qf.  pt. 
7     2     5     11         7  times  I  pint  make  7  pints  ;  2  pints  make 
1    quart ;    then,    7   pints   make  3  quarts,  and 
leaves  1  pint.     Set  down  the  1  pint,  and  carry 
53     2     6     1  the  3  quarts  to  the  product  of  the  next  figure. 

7  times  5  quarts  make  35  quarts,  to  which 
add  the  3  quarts,  which  make  38  quarts;  8  quarts  make  1  peck;  then 
38  quarts  make  4  pecks,  and  leaves  6  quarts.  Set  down  the  6  quarts, 
and  carry  the  4  pecks. 

7  times  2  pecks  make  14  pecks ;  add  the  4  pecks,  and  it  makes  18 
pecks.  4  pecks  make  1  bushel  ;  then,  18  pecks  make  4  bushels,  and 
leaves  2  pecks.     Set  down  the  2  pecks,  and  carry  the  4  bushels. 

7  times  7  bushels  make  49  bushels  ;  add  the  4  bushels,  and  it 
makes  53  bushels,  which  set  down,  and  the  work   is  done. 

bu.  pk.  qt.  pt.  bu.  pk.  qf.  pt. 

y     3     6     1  23     2     5     1 

5  8 


49     3     0     1  189     1     4     0 


In  1  vessel  are  contained  29  bushels  2  pecks  and  5  quarts ;  how 
many  in  9  such  vessels  ?  Ans.  266  bu.  3  pk.  5  qt. 

Observation.  When  the  multiplier  is  more  than  12,  and  a  com- 
posite number,  we  may  multiply  first  by  one  factor,  and  that  product 
by  the  other;  or,  if  the  multiplier  has  more  than  two  simple  factors, 
such  as  72,  which  is  the  product  of   2X4X9,  we    may  multiply 


68  ARITIliMETIC. 


by  the  (^ilT^•rpnt  factors  in  succrssion,  no  matler  which  we  take  first; 
though  it  is  generally  more  expedient  to  take  the  largest  first. 

Orseiivatiox  SKcoxn.  When  the  multiplier  is  not  a  composite 
number,  such  as  23,  we  may  take  the  sum  24  times,  by  multipiyitig 
by  (i,  and  that  product  by  4,  and  then  subtracting  once  the  sum,  which 
will  leave  23. 

If  the  multiplier  be  31,  we  may  multiply  by  10  and  3,  and  after- 
wards add  once  the  number,  &c. 

APrLICATlON. 

1.  What  is  the  cost  of  9  hundred  weight  of  chee-se,  at  1  pound  11 
shillings  and  5  pence  per  hundred  weight  ?  Aris.  £14  2s.  9d. 

2.  A  merchant  has  .3  chests  of  tea,  each  weighing  3  hundred 
weight  2  quarters  and  9  pounds  ;  what  is  the  weight  of  the  whole? 

Ans.  10  cwt.  2  qrs.  27  lbs. 

3.  If  a  man  drink  I  pint  2  giljs  of  ale  per  day  for  29  successive 
days,  what  quantity  will  he  have  drank  in  all? 

Alls.  5  gal.  1  qt.  1^  pt. 

4.  If  a  soldier's  ration  of  bread  be  5  pounds  6  ounces  8  drams  per 
week,  what  will  it  amount  to  in  52  weeks? 

Ans.  2  cwt.  2  qrs.  1  lb.  2  oz. 

5.  A  merchant  bought  9.5  pairs  of  shoes,  at  4  shillings  6  pence  and 
I  quarter  a  [)air ;  how  much  did  he  pay  for  the  whole  ] 

£21  9s.  5Jd. 

6.  A  gentleman  bought  43  silver  spoons,  each  weighing  2  ounces 
14  pennyweights  and  6  grains;  what  was  the  weight  of  the  whole? 

Ans.  9  lbs.  8  oz.  12  dwts.  18  gr. 

7.  How  many  yards  of  cloth  in  35  pieces,  each  containing  27  yards 
3  quarters  2  nails  ?  Ans.  975  yds.  2  qrs.  2  na. 

8.  A  silversmith  has  7  tankards,  each  weighing  3  pounds  4  ounces 
0  pennyweights  and  22  grains;  what  is  the  w^eight  of  the  whole? 

Ans.  23  lbs.  4  oz.  6  dwts.  10  gr. 

9.  If  a  man  can  perform  a  piece  of  work  in  2  years  and  3  months, 
how  long  would  it  take  him  to  perform  5  such  pieces  ? 

Ans.  11  yrs.  3  mos. 

10.  If  a  laborer  dig  a  drain  in  2  weeks  and  3  days,  how  long  a 
time  would  he  require  to  dig  9  such  drains?  Ans.  21  w.  6  d. 

41.  It  is  found  by  oKservation  that  the  sun,  on  an  average,  changes 
his  longitude  0°  59'  8"  33  per  day  ;  how  much  will  he  change  in  10 
days?  how  much  in  30?  how  much  in  300?  how  much  in  365? 

■^   9  51  23  3 


.   I  29  34   9  9 
^^^'   f295  41  39 
J  359  45  40  45 


COMPOUND    DIVISION.  69 


N.  B.  In  this  example,  8''  33  is  8  seconds  and  33  hundreds,  the 
33  being  decimals.  But  we  have  ni)t  yet  exercised  in  decimals  ;  and 
as  thev  increase  and  decrease  on  iht  same  ncaU  as  abstract  numbers, 
we  do  not  tiesilate  to  introduce  them  thus  incidentally. 

In  the  table  of  circular  motion,  the  pupil  will  lind  signs,  degrees, 
and  minutes;  but  the  stilus  arc  no  longer  rdained  in pruc/icul  use ,- 
however,  through  superstition  and  old  custom,  signs  will  long  retain 
their  places  in  common  almanacs. 

12.  The  moon  moves  13*^  10'  35"  in  one  day,  how  many  degrees, 
minutes,  and  seconds,  will  she  move  in  17  days  ? 
4X4-1-1=17 

13°      10'     35 


In  4  days  it  moves 52       42      20 

4 

In  16  days  it  moves 210       49      20 

In  1  day  it  moves 13       10      35 

la  17  days  it  moves 2-i3^     59'     55'' Ans. 

1:3.  The  planet  Jupiter  changes  its  longitude  4'  59"  2  in  one  day 
how  much  will  it  change  its  longitude  in  59  days? 

Ans.  4°  54'  18". 

14.  The  planet  Saturn  changes  its  longitude  2'  0"  91  in  one  day 
how  much  will  it  change  in  3ti5  days  ?  Ans.  12°  15'  37". 


/ 


COMPOUND  DIVISION.. 


(AnT.  31.)  CoMPOUxVD  Division — the  reverse  operation  to  com- 
pound multiplication,  and,  in  its  general  principle,  the  same  as  simple 
division. 

As  in  simple  division,  we  must  divide  the  highest  number  of  the 
dividend  as  a  simple  number,  by  the  divisor ;  and  the  remainder  must 
be  reduced  to  the  next  lower  denomination,  and  the  number  standing 
in  that  denomination  being  added  in,  we  have  a  new  dividend  to  be 
divided  asa  simple  number ;  and  so  on  through  all  the  denominations. 

This  obvious  operation  gives  us  the  following 

RuLK.  Place  the  divisor  on  the  left  of  the  dividend,  as  in  simple 
division. 

1.  Begin  at  the  left  hand,  and  divide  the  number  of  the  highest 
denomination  by  the  divisor^  setting  down  the  quotient  in  its  proper 
place. 

2.  If  there  be  any  remainder  after  this  division,  reduce  it  to  the 
next  lower  denominal/on,  which,  add  to  the  number,  if  any,  belong- 
ing to  that  denomination,  and  divide  the  sum  by  the  devisor. 


70  ARITHMETIC. 

3.  Set  down  cs;uin  this  quotient,  reduce  its  remainder  to  the  next 
h)wer  dtnominalion  again,  and  so  on  through  all  the  denominations 
to  the  last. 

N.  B.  Compound  division  and  compound  multiplication  prove 
each  other. 

EXAMPLES. 

1.  Divide  19  pounds  17  shillings  and  6  pence  equally  among  6 
men. 

Here  we  say,  6  in  19  is  contained  3  times,  and 
1  pound  over,  which,  reduced  to  shillings,  and  17 
added,  make  37  shillings,  which,  divided  by  6,  is 
contained  6  times,  and  1  shilling  over,  which  is  12 
pence,  and  the  6  added  is  18,  which,  divided  by 
6,  gives  3  pence, 
bushels  2    pecks   6  quarts   1   pint   equally  among 

Here  7  into  25  bushels  3  times,  and  4  re- 
mains.    Set  down  the  3. 

Reduce  the  4  bushels  to  pecks,  which  makes 
16  pecks:  add  16  pecks  to  2  pecks,  it  makes 
18  pecks.  Now,  7  into  18  pecks  2  times,  and 
leaves  4.     Set  down  the  2,  &c. 

3.  Divide  821  pounds  17  shillings  and  9^  pence  by  4. 

Ans.  £205  9s.  5d.  3  far. 

4.  Divide  55  pounds  14  shillings  and  |  pence  by  7. 

Ans.  £7  1 9s.  Id.  32  far. 

5.  A  farm  containing  746  acres  3  roods  29  poles,  is  to  be  divided 
equally  between  9  heirs;  what  is  the  share  of  each  1  ^ 

A7is.82  A.  3R.$8^  P. 

6.  Divide  3  acres  of  land  into  8  village  lots  ;  what  number  of  poles 
will  each  lot  contain?  Am^.  60. 

7.  Divide  3  acres  of  land  into  lots,  each  lot  containing  3  roods  ; 
bow  many  lots  will  there  be  ?  Ans.  4. 

8.  Divide  1  acre  1  rood  of  land  into  10  equal  parts;  how  many 
poles  will  each  part  contain  ?  Ans.  20. 

9.  Put  23  bushels  1  peck  of  wheat  into  10  bags ;  how  much  must 
each  bag  hold  ?  Afis.  2  bu.  IJL    pk. 

10.  Divide  10  gallons  and  3  quarts  into  6  equal  portions:  what 
will  each  portion  be]  Ans.  1  gal.  3  quarts  ^  pt. 

(Anr.  32.)  When  the  divisor  is  over  12  and  is  a  composite  number, 
that  is,  one  consisting  of  two  or  more  simple  factors,  we  may  divide 
as  in  short  division,  first  by  one  factor,  and  that  quotient  by  another, 
&c.  But  when  the  number  is  prime,  having  no  simple  factors,  we 
divide  by  the  whole  number  at  once,  after  the  manner  of  long  division. 
Thus. 


£ 

s. 

d. 

6)19 

17 

6 

3 

6 

3 

2.   Divide 

25   I 

persons. 

bu. 

pk. 

r^t.pf 

7)25 

2 

6     1 

3 

2 

5     1 

COMPOUND    DIVISION.  71 


6.  Divide  79  buslicls  1  peck  7  quarts  by  23. 

hu.  pk.qt.  bu.  ph.  qt.pt. 
23)79     17(3171 
69 

—    4X10=40,  and  1  peck  aJded  makes  41,  &c. 
10 
4 

23)41(1  peck. 
23 

18 

23)151(6  quarts. 
138 

13 
2 

23)26(1  pint. 
23 

Rem.  3  pints. 

7.  A  boat  load  of  corn,  containing  4927  bushels  3  pecks,  is  owned 
equally  by  29  persons  ;  what  is  the  share  of  each? 

Ans.  169  bu.  3  pk.  5  qt.  1  pt.,  and  I  pt.  rem. 

8.  Divide  542  pounds  7  shillings  and  10  pence  by  97. 

Ans.  £b  Us.  lOd. 

9.  Divide  123  pounds  II  shillings  2^  pence  by  127. 

Ans.  £0  19s.  5^d. 

10.  Divide  330  hundred  weight  3  quarters  by  14. 

A>u.2ri  cwi.  3  qrs.  18  lbs. 

11.  If  35  pieces  of  cloth,  of  equal  quality,  contain  971  yards  and 

1  quarter,  how  many  yards  in  a  piece  ?  Ann.  27  yd.  3  qr. 

12.  If  259  acres  1   rood   10  rods  of  land  be  divided  into  36  equal 
lots,  how  much  land  will  be  contained  in  a  lot  ? 

Ans.t  A.OTi.  32^  r. 

13.  If  56  pounds  of  butter  cost  4  pounds  18  shillings,  what  is  it 
per  pound?  Ans.  Is.  9d. 

14.  Divide  124  pounds  5  shillings  and  4  pence  into  32  equal  parts. 

Ans.  £3  17s.  8d. 

15.  Divide  336  bushels  3  pecks  4  quarts  by  70. 

Ans.  4  bu.  3  pk.  2  qt. 

16.  Divide  336  bushels  3  pecks  4  quarts,  by  4  bushels  3  pocks  and 

2  quarts.  Ans.lQ. 

Reverse  the  preceding  problems,  taking  the  answers  for  divisors, 


72  ARITHMETIC. 


and  the  former  divisors  will  be  quotients;  but  to  effect  the  division, 
reduce  all  to  the  lowest  denomination  mentioned,  and  divide  as  in  sim- 
ple division. 

If  more  clear,  we  may  enunciate  the  16th  example  thus:  A  farmer 
has  336  bushels  3  pecks  4  quarts  of  wheat  in  his  granary,  which  he 
wishes  to  put  in  casks  to  send  away,  each  cask  containing  4  bushels  3 
pecks  and  3  quarts ;  how  many  casks  will  be  required  ? 

Ans.  70  casks. 

The   operation   is   thus: 

hu.pk.  qt.     hu.  ph.  qi. 
4     a     2  )  336     3     4  ( 
4  4 

19  1347 

8  8 

154  )  10780^70  Ans. 
10780' 


17.  A  certain  secret  association  in  Ireland  paid  a  bill  of  j£4  18s.,  by 
assessing  its  members  Is.  9d.  apiece  ;  how  many  members  were  there  ? 

Alls.  56. 

N.  B.  This  is  but  a  variation  of  example  13  ;  and  in  like  manner 
every  example,  up  to  16,  may  be  varied. 

18.  I  have  a  pitcher  which  holds  2  quarts  and  1  pint;  how  many 
times  can  it  be  filled  from  a  barrel  of  cider,  which  holds  31  gallons  2 
quarts  ?  Ans.  502^ 

19.  How  many  times  will  13  bushels  fill  a  vessel  which  holds  1 
quart  1  pint  and  1  gilH  Ans.  256  times. 

20.  How  many  suits  of  clothes,  each  requiring  3  yards  1  quarter 
and  2  nails,  can  be  made  from  a  roll  of  cloth  containing  27  yards  ? 

Ans.  8. 

MISCELLANEOUS    EXAMPLES, 

Applicable  to  the  preceding  principles. 

1.  At  5  cents  a  quart,  what  will  5  bushels  cost?  Ans.  §8. 

2.  At  3  cents  a  pint,  what  will  3  pecks  and  6  quarts  cost  '\ 

Ans.  90  cents. 

3.  At  5  cents  a  pint,  how  many  gallons  can  be  bought  for 
10  dollars?  Ans.  25. 

N.  B.  As  the  pupil  has  been  instructed  from  the  very  first  of  this 
work,  from  simple  addition,  that  10  mills  make  a  cent,  10  rents  make 
a  dime,  and  10  dimes  make  a  dollar,  and  that  these  denominations  are 
the  same  in  order  as  the  order  of  simple  numbers;  therefore,  the  re- 
duction from  dollars  to  cents  is  to  multiply  by  100;  dollars  into  mills. 


COMPOUND    DIVISION.  73 


multiply  by  1000;  and  reduction  the  other  way  is  to  divide  by  these 
numbers. 

Some  arithmetical  writers  have  treated  Federal  Monei/  as  conipmtnd 
numbers,  ami  have  gone  throui^h  all  the  formality  of  reduction — addi- 
tion, subtraction,  multiplication,  and  division  of  federal  money — the 
same  as  they  do  the  really  compound  numbers,  pounds,  shillings,  and 
pence. 

But,  federal  money  was  purposely  adjusted  to  the  rcuk  nf  simple 
numbers  ,-  and  if  it  is  now  proper  to  treat  these  denominations  as 
compound,  we  must  suppose  the  design  not  accomplished.  Federal 
money  belongs  to  whole  numbers  and  decimal  fractions  ;  and  the  sub- 
ject must  be  incomplete  until  we  pass  decimal  fractioris. 

Example  3,  and  most  of  the  examples  here  inserted,  should  be  done 
by  canceling,  as  explained  in  article  21. 

4.  At  8  cents  a  gill,  how  many  gallons  will  12  dollars  purchase? 

Aris.  4l;. 
.5.   What  will   10  ounces  10  pennyweights  cost,  at  15  cents  a  pen- 
nyweight ?  *  Ans.$-M.50. 

6.  At  20  cents  a  square  rod,  what  will  2  acres  3  roods  of  land 
cost?  Ans.$fi8. 

7.  At  $2  a  square  rod  for  land,  what  must  be  paid  for  a  village  lot 
12  rods  long  and  5-|  rods  wide  ]  Ans.  S132. 

8.  In  11  bars  of  gold,  each  containing  5  pounds  3  ounrcs  2^  pen- 
nyweights,   how  many  grains  ?  Ans.  23:]!^8\ 

9.  At  5  cents  an  ounce,  what  will  10  p.v.inds  i  ounces  of  copper 
cost  ?  A71S.  iS.2(). 

10.  IIov.-  many  kegs,  each  holding  4  traiions  2  .juarts,  can  be  filled 
from  a  liogshead  containing  G3  gallons  ]  Aus.  14. 

1 1.  .'\.t  6^  cents  a  quart,  what  will  6  bushels  atul   1   peck  cost  ( 

Anf,:  $12.r)0. 

12.  How  much  will  4  barrels  of  molasses  cost,  at  4  cents  a  pint. 

Any.  $10.08. 

13.  Divide  2  hours  10  minutes  by  5  minutes  5  seconds. 

Ans.  2^^. 

14.  How  many  times  is  13°  20'  contained  in  360°  ? 

Ans.  27  times. 

15.  If  the  moon  moved  13°  20'  in  1  day,  how  many  days  would  it 
require  to  make  a  revolution  nf  360°  ]  An.^.  27  days. 

leXlf  Jupiter  changed  its  longitude  5  minutes  of  a  degree,  as  seen 
from  the  sun,  bow  many  years,  of  3^60  days  each,  would  it  require  to 
make  a  revolution  ?  -    O       "  Ans.  12. 

17.  At  8  cents  a  pint  for  wine,  how  many  gallons  can  be  bought 
for  40  dollars?  Ans.  6-2^. 

18.  At  10  cents  a  nail,  how  many  yards  of  cloth  can  be  buu^du  for 
16  dollars?  '  J/?.v."]0. 

19.  At  12^  cents  a  quart,  how  many  gallons  can  be  bought  for  12 
dollars?  Ans.  24. 


74 


ARITHMETIC. 


20.  How  many  little  squares,  3  inches  long  and  3  inches  wide,  can 
be  cut  from  a  square  yard  of  paper?  Ans.  144. 

21.  How  many  bottles,  each  containing  1  quart  1  gill,  will  be  re- 
quired   to  draw    off    a    barrel    of    cider   containing    3U    gallons! 

'Ans.  112. 

22.  Divide  421  pounds  14  shillings  and  8  pence  among  3  men,  5 
women,  and  7  boys,  and  give  each  man  double  of  the  sum  given  to  a 
woman,  and  each  woman  3  lanes  the  sum  given  to  a  boy  ;  how  much 


IS   tne   share   ct 


^ 


each  ? 

Ans.  Each  boy  must  have  £10  10s. 
Each  woman,  .  ...  31  12s. 
Each  m.an, 63     53. 


I0|d. 
7id. 
2|d. 


The  art  of  working  the  preceding  problem  consists  in  obtaining  the 
divisor,  which  is  40. 

23.  A  man  bought  a  chaise,  horse,  and  harness,  for  70  pounds.  He 
gave  twice  as  much  for  the  horse  as  for  the  harness,  and  twice  as  much 
for  the  chaise  as  for  the  horse  ;  what  did  he  give  for  each  ] 

A}is.  Harness  £10. 

24.  A  farmer  sold  some  calves  and  some  sheep  for  108  dollars;  the 
calves  at  5  dollars,  the  sheep  at  8  dollars  apiece ;  there  were  twice  as 
many  calves  as  sheep, — what  was  the  num.ber  of  each  sort? 

Aus.  6  sheep  and  12  calves. 

25.  The  planet  Jupiter  changes  its  mean  longitude  4°  54'  18'^  in 
59  days;  how  far  will  it  change  in  one  day  ?  Aus.  4'  59"  2. 

26.  The  moon  is  observed  to  move  over  197°  38'  45"  in  15  days; 
how  far  will  it  move  in  one  day  ?  Ans.  13°  10'  35", 

27.  In  365  days,  the  planet  Saturn  will  change  longitude  12°  15' 


37"  ;  how  much  will  it  change  in  one  day  ? 


A^7is. 


0"  91. 


28.  If  the  apparent. motion  of  the  sun  be  59'  8"  in  one  dav,  how 
many  days  will  it  require  to  make  a  revolution  of  360°? 

Ans.  365|||. 

29.  'I'he  moon  changes  her  longitude  13°  10'  3.5"  in  one  day  ;  how 
many  days,  then,  will  be  required  to  make  one  revolution  ? 

Ans.  27  d.  7  h.  43  m. 

30.  If  Venus  changed  her  longitude,  (as  seen  from  the  sun,)  1° 
36'  per  day,  what,  then,  would  be  the  time  of  her  revolution  1 

Ans.  225  days. 


FRACTIONS, 


SECTION     III. 


FRACTIONS. 

(Art.  33.)     A  part  of  any  one  thing  is  called  ;\  fracjion. 

If  an  api.li',  f)r  instance,  be  divided  into  3  equal  parts,  each  part 
will  he  ont-thlrd,  written  thus,  5. 

If  it  be  divided  into  4  equal  parts,  each  part  will  be  onc-fuurlh, 
wriiten  \. 

!f  divided  into  5  equal  parts,  each  part  will  bo  \  ;  two  of  these 
parts  must  be  written   1. 

i  bu.-,  generally,  a  fraction  must  be  expressed  by  two  numbers  one 
a!)ove  another.  The  lower  number  denotes  the  number  of  equal 
part;^  into  which  the  unit  is  divided. 

The  upper  number  shows  how  many  of  these  equal  parts 
are  taken. 

ileiu-e,  as  the  lower  number  of  a  fraction  denotes  or  decides  the 
den:»ii'iiat;()n,  whether  it  be  thirds,  fourt/iy,  ffths,  or  any  other 
number,  it  is  called  the  denominator. 

'i'l'.e  ur)per  number  is  called  the  nunjf/Y//or,  because  it  shows  the 
number  1  if  parts  taken. 

(A;tT.  31.)  A  fraction  may  be  considered  as  the  result  of  an  im- 
possible division.  Thus,  1,  one  thing,  or  ujiity,  divided  into  5  equal 
parts,  we  write  I  above  and  ^  under  it,  or  1.  'J'hree  times  this  is  |  ; 
or,  we  may  consider  3  divided  by  5,  the  quotient  is   1. 

Hence,  the  denominator  of  a  fraction  may  he  considered  as  a  divisor, 
and  the  numerator  as  a  dividend,  and  the  fraction  itself  as  the  result, 
or  quotient,  to  an  example  in  division. 

When  the  dividend  and  divisor  are  equal,  the  quotient  is  1  ;  that 
i.s,  whiMi  the  numerator  and  denominator  are  equal,  we  have  the  whole 
of  the  thinij;,  the  whole  as  an  aggregate  or  unit. 

(Art.  35.)  When  any  fraction  is  before  us,  as  f,  we  judge  of 
its  value  by  comparing  its  numerator  with  its  denominator  ;  if  the  nu- 
merator is  e^iual  to  the  denominator,  as  we  have  just  observed,  tlie 
value  of  the  fraction  is  1 ;  if  the  numerator  is  nearly  equal  to  the  de- 
nominator, the  value  of  the  fraction  is  nearly  I  ;  if  the  numerator 
is  one-half  of  the  denominator,  the  value  of  the  fraction  i.s  one-half. 
Hence,  1  is  the  same  as  ^. 

6 

(Art.  3fi.)  As  the  value  of  a  fraction  depends  upon  the  relation 
of  the  numerator  to  the  denominator,  and  as  tliis  relation  is  not 
changed  bv  dividing  both  numbefs  by  the  same  divisor,  we  may,  there- 


76  ARITH3IETIC. 


fore,  divide  both  numerator  and  denominator  by  any  number  that 
will  divide  them  without  a  remainder,  and  the  value  of  the  fraction 
will  not   be   changed.* 

EXAMPLES. 

1.  Reduce  Li  to  its  lowest  terras.  Ans.^. 

-  4 

Here,  it  is  evident  that  we  can  divide  both  numerator  and  denomi- 
nator by  6,  making  the  fraction  |,  which  is  equal  to  li. 

2.  Reduce   il  to  its  lowest  terms.  Ans.  f. 

5  6 

3.  Reduce  _3JL  to  its  lowest  terms.  An^.  ^. 

When  the  terms  of  the  fraction  are  large,  we  may  not  bring  it  to  its 
lowest  terms  by  one  division  only,  but  we  may  divide  the  quotients  ob- 
tained continually  until  no  number  greater  than  1  will  divide  both 
of  them  \Nithout  a  remainder. 

4.  Reduce  Ll^  to  its  lowest  terms. 

7  3  0 

5.  Reduce  1?.^  to  its  lowest  terras. 

2  0  4 

6.  Reduce  i?.  1  to  its  lowest  terms. 

6  3  0 

7.  Reduce  J^^-  to  its  lowest  terms. 

8.  Reduce  119.  to  its  lowest  terms. 

8  0  0 

9.  Reduce  iii  to  its  lowest  terras. 

4  9  2 

10.  Reduce  15?-  to  its  lowest  terms. 

11,  Reduce  -1^-  to  its  lowest  terms. 


*  As  a  guide  to  the  studem  to  find  suitable  divisors  for  these  reduclionSjIet 
him  observe, 

1st.  That  any  number  ending  with  an  even  number  or  a  cipher,  can  be 
divided  by  -2. 

2d.  Any  number  ending  virith  5  or  0.  is  divisible  by  5. 

3J,  If  tiie  right  hand  place  of  any  number  be  0.  the  whole  is  div;s'l>le  by 
10;  if  there  be  two  ciphers,  it  is  divisible  by  100;  if  three  ciphers,  by  lOCO. 
and  so  on.  which  is  only  cutting  ofl"  tho?e  ciphers. 

4th.  If  the  two  right  hand  figures  of  any  number  be  divisible  liy  4,  the 
whole  is  divisible  by  4:  and  if  the  three  right  hand  figures  be  divisible  by 
8,  the  whole  is  divisible  by  8.  and  so  on. 

5th.  If  the  sum  of  the  dig  ts  in  any  number  be  divisible  by  3  or  by  9,  the 
whole  is  divisible  by  3  or  by  9. 

6th  If  the  ri?ht  hand  digit  be  even,  and  the  sum  of  all  the  d.g;is  be  divisi- 
ble by  6,  then  the  whole  is  divisible  by  6. 

7th.  A  number  is  divisible  by  11,  when  the  sum  of  the  1st.  3d.  5ih.  kc.  or 
all  the  odd  places,  is  equal  to'  the  sum  of  the  2d,  4th,  6th,  &c.,  or  of  all  the 
even  places  of  digits. 

Sih.  If  a  number  cannot  be  divided  by  some  cpiantity  less  than  the  square 
root  of  the  same,  that  number  is  a  prime,  or  cannot  be  divided  by  any 
number  whatever.  ,    „  ^        ^   ■      , 

9th.  All  prim-.'  numbers,  except  2  and  5,  have  either  1.  3.  /.  or  9.  in  the 
Dlace  of  units;  and  all  other  numbers 're  composite,  or  can  be  divided. 


Ans 

i. 

Ans 

i 

Ans. 

.5, 

6* 

Ans.  2 

J  ^ 

Ans.  ^ 

o' 

Ans. 

h 

Ans.). 

'J  ^ 
i' 

Ans.  ^%_ 

r 

FRACTIONS.  77 


I     12.    Reduce  3.2  i P.  to  its  lowest  tcriii:^.  ,-l//.v.  J,. 

tj  4  8  II  - 

13.  Reduce  J.JJ'J'J^-  to  ils  lowest  leriiis.  Ajis.  1. 

6  7   18  5  1)  « 

14.  Reduce  -JJJ'J'jyL  to  its  lowest  terms.  A7is.  }. 

15.  Reduce  _6_7_8AP_  to  its  lowest  tciins.  Ans.  J_ 

0  7  8  6  0  0  in' 

16.  Reduce  -A<L'^j'-  to  its  lowest  terms.  Ans.  'II. 

10  0  0  0  S  C  5 

(AuT.  37.)  In  place  of  dividing  by  small  common  divisors,  in 
succession,  we  may  lind  the  greatest  common  divisor  at  once,  and 
divide   by  it. 

RuLK. /Tc)  Jhid  the  greatest  common  divisor  bctivcen  two 
numbers  .• 

Divide  the  greater  nutnhcr  by  the  smaller,  and  this  divisor  hi/  the 
remainder,  and  thus  continue  dividing  the  last  divisor  hi/  the  last 
remainder,  till  nothing  remains.  The  divisor  last  used  will  be  the 
number  required. 

What  is  the  greatest  common  measure  of  918  and  1944? 

918)1944(2 
1836 

108)918(8 
SG4 

64)108(2 
108 


The  truth  of  the  rule  in  this  problem  will  he  discovered  by  retracing 
the  above  operation,  as  follows:  feince  54  (the  last  divisor)  measures 
108,  it  also  measures  8X  l08-|-64,  or  918.  Again,  since  .54  measures 
108  and  918,  it  also  measures  2X^18-1-108.  or  1944.  Therefore,  54 
measures  both  918  and  1944.  It  is,  also,  \\\e  greatei-t  common  mea- 
sure ;  for,  suppose  there  be  a  greater —  then,  since  the  gre:iter  mea- 
sures 918  and  1944,  it  also  measures  the  remainder,  108  ;  and,  since 
it  measures  108  and  918,  it  also  measures  the  rcmaiiuier,  54,  that  is, 
the  greater  measures  the  less,  wliich  is  absurd. 

EXAMPLES. 

1.  V\'hat  is  the  greatest  common  measure  of  the  numbers  3;::?  and 
425  1  Ans.  17. 

2.  What  is  the  greatest  common  measure  of  2310  and  462G  1 

Ans.  G. 

3.  Reduce  2.?.i  to  its  lowest  terms. 

4  2  .s 

4.  Reduce  r^'llH.  to  its  lowest  terms. 

4  0  2  6 


g2 


78                                                    ARITHMETIC. 

5.   Reduce  ^6_;  3    to  its  lowest  terms. 

5  9  4  0 

An..  1». 

6.  Reduce  2.5.  co  to  its  lowest  terais. 

5  ■•)  4  0 

7.  Reduce  2  7  5^  to  its  lowest  terms. 

440 

Ans.  f . 

8.   Reduce  -Yy  g-  to  its  lowest  terms. 

Ans.  ^=^.. 

9.  Reduce  JJ>J^  to  its  lowest  terms. 
1057 

Ans.  4. 

(AuT.  c58.)      We  have  thus  far  been  s 
in  their  simplest  forms ;  but  we  have  also 
pound  Fractions,  Mixed  Numbers,  and 

peaking  ol 
Improper 
Complex 

'  proper  fractions, 
Fractions,  Corn- 
Fractions. 

An  improper  fraction    has     a     numerator      greater      than     its 
denominator,  as  t^^  ^^  Sec. 

A  compound  fraction  is  a  fraction  of  a 

fraction,  as  ^  of  f . 

A  mixed  number  is  a  whole  number  and  a  fraction  standing  in  one 
sum,  as7|-,  11^,  19|,  &c. 

A  complex  fraction  is,  where  one  or  both  of  the  ie7'ms  of  the  frac- 
tion are  fractional,  as 

2       71     2| 

3L,    9,      51. 

By  an  improper  fraction,    we  understand  that  the  value  of  the  ex- 
pression is  more  than  a  unit  in   value:  thus,  1.     As  3.  make  1,  1 

must  equal  2^. 

Hence,   we  may  reduce  improper  fractions  to  mixed  numbers,  by 
dividing  the  numerator  by  the  denominator,  and  taking  the  remain- 
der for^the  numerator  of  the  proper  fraction. 

EXAMPLES. 

1.  Reduce  '_?-7  to  its  proper  terms. 

Ans.  11_5_. 

2.  Reduce  -i  1 1  to  its  proper  terms. 

Ans.  153f 

.3.  Reduce  ^J/    to  its  proper  terms. 

Ans.  2^-. 

4.  Reduce  1  £1  to  its  proper  terms. 

Ans.  301. 

5.  Reduce  ^_6_i  to  its  proper  terms. 

Ans.  56 ^!L, 

6.  Reduce  '  |5  s  to  its  proper  terms. 

Ans.  54^1 . 

7.  Reduce  '-P  to  its  proper  terms. 

Ans.  1|. 

8.  Reduce  >  1."  to  its  proper  terms. 

Ans.  23|. 

9.  Reduce  '^J^-?  to  its  proper  terms. 

An-.  41. 

10.  Reduce  "isi  to  its  proper  terras. 

Ans.  430^. 

VULGAR    FRACTIONS.  79 

(AiiT.  39.)  Whole  numbers  may  be  put  under  a  fractional  form, 
by  placing  unity  under  them.     Thus,  i.  is  manifestly  4,  ^  is  9,   Y 

is  44,  &c. 

Now,  by  Article  36,  the  value  of  any  fraction  is  not  changed  by 
dividing  both  numerator  and  denominator  by  the  same  number.  In- 
versely, then,  we  shall  not  alter  the  value  of  any  fraction  by  multiply- 
ing both  its  terms  by  the  same  number.  Hence,  we  can  reduce  any 
number  into  halves,  thirds,  fourths,  or  any  other  number  of  equal 
parts,  by  first  placing  unity  under  it,  and  then  multiplying  the  nume- 
rator and  denominator  by  the  parts  required,  thus  :  Reduce  (5  units  to 
fifths.  First,  1 ;  then  multiply  both  terms  by  5,  and  we  have  3_o 
Answer.  Again  :  Reduce  62  to  fifths.  The  result  is  obviously  3_2 
—  and  from  this  we  may  draw  a  rule  to  reduce  mixed  numbers  to  im- 
proper fractions. 

Rule.  Multiply  the  whole  number  by  the  denominator  of  the 
fraction,  and  add  in  the  numerator  for  a  new  numerator,  and  set 
the  denomitiator  under  it. 

EXAMPLES. 

1.  Reduce  4^  to  an  improper  fraction.  Ans.  1, 

2.  Reduce  711  to  an  improper  fraction.  Ans.  •'^.a. 

3.  Reduce  111  to  an  improper  fraction.  Ajis.  i|. 

4.  Reduce  37§  to  an  improper  fraction.  Ans.  "-^  . 

53 

5.  Reduce  5i  to  sixteenths:  thus,  --X  i-f  =  f f ,  ^«*- 


6.  Reduce  21  to  sevenths.  Ans. 


14    7) 

7 


3_6  7 
4 


7.  Reduce  12J_  to  an  improper  fraction.  An 

8.  Reduce  341|  to  an  improper  fraction.  Ans. 

9.  Reduce  241  to  tenths.  Aiis.  2_y  • 

72^ 

10.  Reduce  24 1,  to  thirds.  Atis.  -g- 

11.  Reduce  67-^-  to  an  improper  fraction.  Ans.  7_4_' . 

12.  Reduce  131:|  to  an  improper  fraction.  Ans.  3|5  ^ 

13.  Reduce  211  to  an  improper  fraction.  Ans.  1 1^ 

If   more  examples  are  desired,  the  pupil  can  reverse  those  under 
Article  38. 


80  ARITHMETIC. 


CO.MrOUND  FRACTIONS, 

(Art.  40.)  We  have  before  dofined  Conipountl  Fractions  to  be 
fractions  of  other  fractions,  such  as  ^  of  ^,  £.  of  ^,  &c. 

When  a  compound  fraction  is  reduced  to  its  simple  value,  it  is  evi- 
dent that  that  simple  value  must  he  less  than  either  fraction  mentioned, 
because  it  must  be  part  of  a  per/. 

CompoHiid,  in  arithmetic,  always  means  a  product ;  and,  to  reduce 
compound  fractions  to  single  fractions,  we  must  multiply  them  toge- 
ther. Hmce,  (he  same  problems  may  appear  under  multiplicaiiun 
of  fractions  and  under  ccmpound  fraclioJis. 

To  find  the  value  of  a  comriound  fraction,  let  us  observe  that  ^  of  -f 
is  certainly  5  ;  and  we  can  obtain  the  same  result  by  nuiltipiying  l)oth 
numerators  together  for  a  new  numerator,  and  the  denominators  for 
a  new  denominator,  and  then  reducing  down  ;  or  we  may  cancel  the 
twos  in  the  numerator  and  denominator. 

Again  :  what  is  the  value  of  ^  of  -f  of  ?.  ? 

One-half  of  f  is  certainly  ^  ;  then  f  of  ?.  is  certainly  less  than  2, 
and  a  fraction  is  made  less  by  increasing  its  denominator.  If  we 
double  the  denominator  only,  the  fraction  is  one-half  its  former  value  ; 
if  we  increase  the  denominator  by  3  times  its  former  value,  the  frac- 
tion will  he  one-third  of  its  former  value,  &c.  Hence,  J^^  is  the  value 
of  5  of  -f  of  ^.  From  these  observations  we  draw  the  following  rule 
for  the  reduction  of  compound  fractions  to  single  ones. 

Rule.  Reduce  mixed  numbers  In  improper  fractions,  and 
place  a  unit  under  whole  members  to  put  them  in  a  frac- 
tional form. 

Then,  mxiliiplii  all  the  numerators  together  for  a  new  numera- 
tor, and  all  the  denominators  for  a  new  denominator,  and  reduce  to 
Us  lowest  terms.  But,  skillful  operators  reduce  by  canceling,  while 
performing  the  general  operation. 

EXAMPLKS. 

1.  Reduce  ^  of  'i  of  §  lo  a  simple  fraction.  Ans.  _^_. 

2.  Reduce  -J  of  1  of  l*.  of  t  to  a  simple  fraction.  Ans.  2L, 

3.  Reduce  3  of  1  of  _?_  of  ^  to  its  simple  value.  ^4^5.  _s_ 
A.  Reduce  2^  of  I5  of  £  of  |  to  its  simple  value.  Ans.  2. 

5.  Reduce   ^   of  i*   of   J    of    ll    of  8  to  its  simple   value. 

Ans.  6_5_ 

6.  Reduce  ^  of  ^  of  J  of  i  of  A  of  il  of  10  to  its  simple  value. 

Ans.  \l 


VULGAR    FRACTIONS.  81 


(AiiT.  41.)  To  invcstig;itc  and  ionn  a  rule  to  multiply  a  fraction 
by  a  whole  jiuniber,  we  can  take  some  simple  fraction,  as  ^.  and  if  it 
he  miiltipliml  by  2,  the  result  must  evidently  be  ?  ;  and  if  it  be  mul- 
tiplied by  3,  the  result  must  be  | ;  if  by  5,  the  result  must  be  s,  &c. 
Also,  take  ^,  and  multiply  it  by  4  ;  the  result  will  be  1,  or  ^.  Also, 
f  multiplied  by  2  will  give  i  ;  2  multiplied  by  3  is  _^,  or  ■§ ;  and  from 
these  observations,  to  multiply  a  fraction  by  a  whole  number,  we  de- 
duce the  foilnwing 

RuLK.  Mulfi'p/i/  the  numerator  by  the  whole  number  ,■  or,  when 
you  can,  divide  ihe  denominator  by  the  whole  number. 

EXAMPLES, 

1.  Multiply  ii  by  5.  Ans.  y=2^- 

2.  Multiply  ^  by  3.  Ans.  ~=^h- 

3.  Multiuly  Li  by  4,  Ans.  «l=3ll. 

4.  Multiply  _!'_  by  100.  Ans.  9-^=642. 

5.  Multiply  ^1  by  IS.  Ans.  1\. 

6.  Multiply  1^3  by  19.  Ans.  5^  =  . 

7.  Multiply  1  by  24.  Ans.  56. 

8.  Multiply  11  by  105.  Ans.  85. 

9.  Multiply  3  by  63.  Ans.  27. 

10.  Multiply  1  by  40.  Ans.  25. 

11.  Multiply  _"-  by  11.  Ans.  7. 

To  multiply  a  whole  number  by  a  fraction  is  the  same  as  to  multi- 
ply a  fraction  by  a  whole  number ;  for,  when  two  numbers  are  multi- 
plied tot^ether,  it  is  indifferent  which  we  call  the  n)ultiplier,  or  which 
the  multiplicand.  (See  Art.  11.)  That  is,  in  the  last  example,  ^- 
multiplied  by  U,  is  the  same  as  U  multiplied  by  _'_  ;  and  so  of  the 
other  examples. 

(Art.  42.)  If  we  mulliply  a  fi-adion  by  its  denominator,  it 
luill  produce  the  iiumerator  for  a  product. 

EXAMPLES. 

1.  Multiply  1  by  7.    By  the  above  rule,  the  product  must  be  ^, 

or  3.  .       -,  A 

2.  Multiply  11  by  19.  Ans.  14. 

3.  Multiply  5  by  9.  Ans.  5. 


ARITHMETIC. 


4.  Multiply  3i  by  3.  Ans.  10. 

5.  Multiply  7i  by  4.  Ans.  29. 

6.  Multiply  6 1  by  5.  ^?is.  33. 

7.  Multiply  1^  by  17.  Aris.  11. 

8.  Multiply  ^^  by  29.  ^ns.  19. 

9.  Multiply  33  by  7.  Ans.  24. 

10.  Multiply  41.^  by  17.  A7is.  79. 

11.  Multiply  15- by  21.  Ans.  13. 

N.  B.  Let  the  pupil  remember  this  article,  when  he  comes  to 
clearing  equations  of  fractions  in  Algebra. 

(Akt.  43.)  We  have  already  defined  Complex  Fractions  to  be 
such  as  have  fractions  in  the  numerator  or  denominator,  or  in  both,  as 

— '-.     Now,  if  we  multiply  both  numerator  and  denominator  of  this 

fraction  by  2,  we  shall  have  J  ,  a  simple  fraction. 

But,  why  did  we  multiply  by  2,  in  preference  to  any  other  number  ? 
Ans.  Because  it  was  the  denominator  in  the  fractional  part  of  the 
numerator. 

3  94 

Again  :  —  is  the  same  as  1,  or  -f.     Also,  —  is  a  complex  fraction, 
4]^  ^  8g 

and  we  can  clear  its  numerator  of  the  fraction  by  rauhiplying  it  by  3, 
and  we  can  clear  its  denominator  by  multiplying  it  by  8.  Hence, 
we  can  banish  fractions  from  both  numerators  b}'  multiplying  by  3 
and  then  by  8,  or  multiplying  both  numerator  and  denominator  by  24 
at  once. 

Thus:  9-1X24=232;  8|X24=20I.  Therefore,  232  is  the 
equivalent  simple  fraction. 

It  is  now  apparent  that  we  can  change  complex  fractions  to  simple 
fractions,  by  the  following 

Rule.  Multiply  both  numerator  and  denominator  by  the  denomi- 
nators of  the  partial  fractions,  or  by  their  product,  or  by  their  least 
common  multiple. 

EXAMPLES. 
2i 

1.  Reduce  — ^  to  a  simple  fraction.  Ans.  ll. 

41  *8 

7 

5 

2.  Reduce  ^rr  to  a  simple  fraction.  Ans.  1  =  11, 

.0  o  5  5 


VULGAR    FRACTIONS.  83 


3.  Reduce  -?  to  a  simple  fraction.  Ans.  2.1. 


i    „.   ^^  „,   4| 


4,  Reduce     -  of    o,   of   -  ^  (o  a  simple  fraction. 

1 3  **3  4 


3  4 


A71S.  4_8^. 


5.  Reduce  ~~   to  a  simiile  fraction.  A)is.  ,"/_. 

1 9  _  I  y  o 

1  24  3^ 

6.  Reduce  —  of  --  of   —  to  a  simple  fraction.        Ans.  tl. 

03.  2  2  ^  ^ 

7.  Reduce    -^  of    ^,   to    a  simple  fraction.  Ans.  -k. 

8.  Reduce  ;;:rj  of  •§  to  a  simple  fraction.  Atis.  _«_. 

If  we  divide  i  by  2,  that  is,  one-half  cut  into  2  equal  parts,  each 
part  will  be  i.  If  we  divide  ^  by  3,  or  in  other  words,  cut  ^  into  3 
equal  parts,  each  part  must  be  i.  But,  we  can  obtain  these  results 
mechanically,  by  multiplying  the  denominator  of  the  fraciiun  by 
the  divisor. 

If  we  were  required  to  divide  2.  into  3  equal  parts,  each  part  will  be 
;..  In  this  case,  then,  we  can  divide  the  fraction  by  dividing  the 
numerator,  because  it  is  susceptible  of  being  divided  into  the  parts 
required. 

Therefore,  to  divide  a  fraction  by  a  whole  number,  multiply  the 
denominator  of  the  fraction  by  the  whole  number,  and  place  the 
numerator  over  the  product.  Or,  when  you  can,  divide  the  nume- 
rator by  the  whole  number,  and  let  the  denominator  remain  unchanged. 

EXAMPLES. 

1.  Divide  I  by  3.  Ans.  i. 

2.  Divide  2  by  9.  Atis.  _2_.. 

3.  Divide  |.  by  5.  Ans.  ^. 

4.  Divide  11  by  13.  Ans.  J 


1  9 


.5.  Divide  _»_  by  8.  Ajis.  _"_. 

6.  Divide  LL  by  7.  Ans.  JJ^. 

2  1-  14  7 


8  4  ARITHMETIC. 


7.  Divide  LI  by  34.  Ans.  J„. 

2  1-'  42 

8.  Divide  ^1  by  15.  .  /         Ans.  ^. 
Tiiis  is  the  reverse  operation  of  Art.  41. 

[N.  B.  Let  pupils  omit  the  following  Article  the  first  time  going 
I  h  rough.]  .'- 

CONTINUED  FRACTIONS. 

(Art.  44.)  We  have  a  clearer  conception  of  the  value  of  a  frac- 
tion when  its  numerator  is  unity,  than  when  it  is  in  any  other  form. 
For  in-stance,  we  have  a  clearer  understanding  of  the  fraction  i  than  of 
-?^J-,  although  they  differ  in  value  but  veri/  little. 

As  the  value  of  a  fraction  is  not  changed  by  dividing  both  nu- 
morator  and  denominator  by  the  same  number,  (Art.  36,)  we  can 
always  change  the  numerator  to  unity,  by  dividing  both  terms  of  the 
fraction  by  the  numerator;  but  this  will  make  the  denominator  a 
mixed  number,*  and  the  fraction  a  complex  fraction. 

If  we  take  the  fraction  -||^,  and  divide  both  numerator  and  deno- 
minator by  287,  we  have  — ,  which  shows  that   the  value  of  the 

fraction  is  between  \  and  \,  and  5  is  its  first  approximate  value. 
If  we  take  -J.|j,  and  divide  its  numerator  and  denominator  by  131, 

we  then  have  ;— -— ,  and  the  original  fraction  becomes 
1 

If  we  continue  to  operate  on  the  fraction  ^J^ ,  until  we  find  the 
last  fraction  with  unity  for  a  numerator,  the  original  fraction,  ||^, 
will  take  the  following  form,  which  is  a  continued  fraction. 

1 

2-t-l_ 
5-t-l__ 

6 

*That  is:  \n  case  the  numerator  and  denominator  are  prime  to  each 
other,  as  il  is  supposed  ihev  are.  If  they  are  not,  the  fraction  must  be  re- 
<luced  bv  Article  36. 


FRACTIONS.  S;> 


If  we  noglect  tlie  fractional  paij  of  the 2ad  denominator,  ihc  second 
apjToximatc  fraction  will  be 

The   first  approximate    value  of  the  original  fraction,  2  "t,    is,  as 

we  have  just  observed,  ^.     The  second  apjiroxiinatt-  value  is  — 

2 
I 
The  third  approximate  value  is  3T71 

5 
The  fourth  approximate  value  is  

3-fI_ 

2-fl__ 
5-1-1 
4 
And  the  fifth  and  last  approximate  value  is  the  fraction  itself 

Let  it  be  observed,  that  the  principle  here  involved  is  all  encom- 
passed in  Article  36,  and  that  we  reiluce  fractions  to  this  shape  by 
division  ;  and,  of  course,  to  reduce  thorn  back  again,  all  we  would 
have  to  do  is  to  inspect  the  work,  and  take  the  reverse  operation. 
For  example,  what  simjile  fraction  expresses  the  value  of  the  third 
a[)proximate  value  of  the  traction  under  consideration] 

We  must  commence  with  the  last  denominator,  1,  and  multiply 
the  numerator  and  denominator  of  the  fraction  —7—:  hy  5,  which  gives 

5 

1 

_5_;  then,  our  approximate  fraction  becomes =  ?-l. 

11'  '  ^'^  3_5_      3  8* 

In  this  manner  we  shall  find  the  approximate  values  of  the  fraction 

fnhpi      2       11         4fi        287 

By  inspection,  we  perceive  that  a  continued  fraction  may  be  defined 

thus : 

A  continued  fraction  is  one  that  has  a  whole  number  and  a  frac- 
tion for  it'^  denominator,  which  fraction  has  ahio  a  whole  number 
and  a  fraction  for  its  denominatur,  aiid  so  on,  as  far  as  the  frac- 
tion may  extend;  ai}d  each  fraction  must  have  unity  fur  its 
numerator. 


86 


ARITHMETIC. 


The  partial  fractions  which  compose  any  continued  fraction,  are 
called  hit egrut  fractious,  because  each  of  their  numerators  must  con- 
sist of  a  single  integer.  The  integral  fractions  which  compose  the 
continued  fractions,  which  we  now  have  under  inspection,  are  5,  5, 
L,  1,  1.  But,  we  have  just  seen  that  the  approximating  fractions 
are  \,  2    n     4_6_    28.7 

By  investigation,  we  find  that  the  approximating  fractions  can  be 
drawn  from  the  integral  fractions,  by  the  following 

Rule.  1.  The  first  approximating  fraction  is  the  same  as  the 
first  integral  fraction. 

2.  The  second  approximating  fraction  has  the  denominator  of 
the  second  integral  fraction  fur  its  numerator,  and  the  product  of 
the  first  two  integral  denominators,  plus  1,  for  its  denominator. 

3.  The  third  approximating  fraction  may  be  fanned  by  multi- 
plying the  terms  of  the  second  approximating  fraction  by  the  de- 
nominator of  the  third  integral  fraction,  and  to  the  products  add 
the  terms  of  the  first  approximating  fraction. 

And  thus,  in  general  terms,  to  form  any  approximating  fraction 
after  the  second, 

Multiply  the  terms  of  any  approximating  fraction  by  the  deno- 
minator of  the  following  integral  fraction,  and  to  the  products 
add  the  corresponding  terms  of  the  preceding  approximating  frac- 
tion, and  we  shall  have  the  terms  of  the  next  succeeding  approxi- 
mating fraction. 


1.  Reduce  I2  i_  to  a  continued  fraction  :  find  its  integral  fractions, 
and  its  approximating  fractions. 

Ans.    ^,    I,      \,        1,        L.  integral  fractions. 

5  '  G  ' 

i      3^    Ll^    --^-^    -r^^i  approximating  fractioiis.      ^ 

2.  Reduce  the  fraction  _Y_  into  a  continued  fraction,  and  find  its 
integral  and  its  approximating  fractions. 

Ans.   ^,    5,     ^,       5,        1,       I,  integral  fractions. 


APPLICATION. 

3.  The  circumference  of  any  circle  is  three  times  its  dinnip*'r, 
plus  the  fraction  -.yj-^JL.  Reduce  this  fraction  to  a  continued  frac- 
tion, and  find  its  first  three  integral  fractions,  and  its  first  three  ap- 
proximating fractions. 

Alls.  |,     _?_,        1.,  integral  fractions. 


Y'   rVe'   TtV'  approximating  fractions. 


VULGAR    FRACTIONS.  87 

By  putting  the  integral  3  to  these  fractions,  and  reducing  to 
improper  fractions,  we  have  2_2^  y|-,  liA,  as  the  approximating 
ratios  between  the  circumference  and  the  diameter  of  a  circle. 

4.  The  solar  year  is  5  hours  48  minutes  49  seconds  above  365 
da3's:  wiuil  part  of  24  hours,  is  5  hours  48  minutes  and  49  seconds  ? 

To  answer  this  question,  we  llrst  reduce  both  of  these  quantities  to 
seconds:  fi  hours  48  minutes  49  seconds  =20929  seconds;  24 
hours  =8(5100  seconds.  Then,  the  proportional  part  that  n  hours 
48  minutes  and  49  seconds  is  of  24  hours  is  expressed  by  the  fraction 

2  0  3  L'  n 
8  6  4  00* 

Reduce  this  fraction  to  a  continued  fraction,  and  find  its  integral 
fractions  and  its  approximating  fractions. 

j„r  X    1    1.  1.  J_.  1.  L.  J_.  are  its  integral  fractions. 

And  its  first  six  approximating  fractions  are 

5i     29'    33'     128'     I61'2704* 

These  fractions  show  that  in  every  four  years  we  must  have  one 
leap  year,  to  make  the  sun  cross  the  equator  on  the  same  day  of  the 
month,  and  this  will  be  only  an  approximate  correction.  Seven  leap 
years  in  29  years  would  be  better,  or  8  in  33,  or  31  leap  years  in  128 
consecutive  years,  would  be  a  very  near  correction,  &c. 

5.  The  planet  Mercury  performs  a  revolution  round  the  sun  in  87 
days  23  hours  and  15  minutes,  nearly,  or  126676  minutes.  The 
earth  revolves  round  the  sun  in  365  days  5  hours  and  49  minutes, 
nearly,  or  in  525949  minutes. 

Hence,  126676  revolutions  of  the  earth  correspond  in  time  to 
525949  revolutions  of  Mercury  ;  for,  525949  X  1-6676  gives  the  same 
product  as  126676X525949.  From  this,  it  appears  that  if  the  earth 
and  Mercury  are  in  any  particular  position  in  respect  to  longitude,  as 
seen  from  the  sun,  they  cannot  both  be  in  the  same  position  again,  at 
the  same  time,  until  after  an  elapse  of  126676  years;  but  they  can 
approximate  to  the  same  position  at  different  intervals,  which  can  be 
determined  hy  making  a  fraction  of  the  numbers  126676  and  525949, 
thus   1-116  7^       Reduce  this  fraction  to  a  continued  fraction,  and  find 

'525949  .  .  ^  . 

its  integral  fractions  and  its  approxnnatmg  tractions. 

Am.  '     I    i    L    t    1    1    i.  J-.  -'-.  1,  are  the  integral  frac- 

-^4'    IT'    T'    T'    2'     1'     1'    6»    6  0'     1'    3'  " 

tions,  and  its  approximate  fractions  are 

1  n  7  13  3  3  ^6  T_9_       J5_2_0_        J}J_2jJ}__       J^JJTLPJL 

4'     IjI'      l^y"'      51'     T3T'      T;i   I'      328'      2159»      129808'      13  2027' 

1 2 Ifi 76 

52  5  i*  4  9  * 

The  first  fraction  shows  that  1  revolution  of  the  earth  has  an  ap- 
proximate value  to  4  revolutions  of  Mercury.  The  next  fraction  shows 
that  6  revolutions  of  the  earth,  (or  6  years,)  has  a  nearer  approximate 
value  to  25  revolutions  of  Mercury,  and  so  on,  as  the  reader  can  see 
by  inspection. 


88  ARITHMETIC. 


N.  B.  We  are  aware  that  ihe  applicaiioii  of  contnuied  fraciioiis  does 
not  belong  to  arithmetic,  for  it  involves  associations  of  ideas  entirely  too 
h;gh  ;  but  we  know  from  experience  thai  pupils  in  general  will  not  take 
sufficient  interest  in  abstract  continued  //actions  to  ui'.(h-;rstand  them,  unless 
they  are  convinced  of  their  practical  utility;  and  this  is  an  apology  lor  giv- 
ing the  mailer  iubsequenl  to  example  2. 


REDUCTION  OF  FRACTIONS. 

(AnT.  45.)  We  have,  thus  far,  treated  fracuon.s  in  the  abstract. 
We  shall  now  begin  to  apply  them  to  particular  things,  and  change 
them  from  »>ne  scale  of  weight  or  measure  to  another.  But,  we  shall 
be  best  understood  by 

EXAMPLES. 

1.  Reduce  |  hundred  weight  to  the  fraction  of  a  pound.  This  is 
ihe  same  as  asking  the  value  of  i  of  » 12  pounds,  which,  by  Article 
40,  must  be  A)is.  ^±^. 

2.  What  is  the  value  of  |  of  1  pound  Troy  ?  that  is,  what  is  the 
value  of  I  of  »_2  ounces  ?  Ans.  7  oz.  4  dwts. 

From  the  foregoing  observations  we  deduce  the  following 

Rule.  To  find  the  value  of  a  fraction  in  parts  of  its  integer  of 
lower  denomination. 

Multiply  the  numerator  by  the  parts  in  the  next  inferior  denom- 
ination, and  divide  the  product  by  the  denominator,  and  multiply 
the  remainder,  if  ariy,  by  the  parts  in  the  next  denomination,  arid 
so  on.  The  quotients  placed  in.  cn-der,  will  express  the  value  of  the 
fraction. 

EXAMPLES. 

1.  What  is  the  value  of  |.  of  an  Ell  English,  in  integers  of  lower 
order?  Ans.  3  qr.  2^  na. 

2.  What  is  the  value  of  |  of  an  acre,  in  integers  of  lower' order? 

Ans.  3  R.  131  P. 

3.  What  is  the  value  of  I  of  a  hogshead  ?  Ans.  49  gals. 

4.  What  is  the  value  of  2  of  3  of  one  hour?         Ans.  30  min. 

5.  What  is  the  value  of  -?^  of  a  day  in  integers? 

Ans.  3  h.  12  min. 

6.  Reduce  j  of  a  dollar,  or  100  cents,  to  its  integer  value. 

Ans.  12^  cts. 

7.  Reduce  |o  of  a  hogshead  to  its  integer  value.     Ans.  50  gals. 

8.  Reduce  '  |.  of  a  day  to  its  integer  value. 

Ans.  16  h.  36  m.  55J5_  sec. 

9.  Reduce  ||  of  a  hogshead  to  its  proper  quantity. 

Ans.  33  gal.  3  qt.  I  pt.  11. 


REDUCTION    OF    FRACTIONS.  89 

10.   Reduce  1^,-  of  a  £  to  its  proper  quantit}-.     Ans.  18  s.  4  d. 

A  fraction  may  be  so  small,  that  is,  the  denominator  so  large,  in 
relation  to  the  numerator,  that  we  can  obtain  no  integer  of  the  wfe- 
rior  decree.     Observe  the  following 


EXAMPLES. 

11.  Reduce  -----  of  a  hogshead  to  the  fraction  of  a  pint. 

Ans.  12. 

8  S 

12.  Reduce  -  J__  of  a  day  to  the  fraction  of  a  minute. 

15  8  4  •' 

Ans.  I'L. 

13.  Reduce ." of  an  acre  to  the  fraction  of  a  rood. 

14  4  0 

Ans.  ^1-. 

^8  0 

14.  Reduce  — 1 of  a  mile  to  the  fraction  of  a  rod. 

10  5  0  0 

Ans.  _5_, 

3  3 

15.  Reduce  ^2_  of  a  £  to  the  fraction  of  a  shilling.        Atis.  ^  s. 

16.  Reduce  j^^-^  of  a  hundred  weight  to  the  fraction  of  a  pound. 

Ans.  A. 

7 

(Art.  46.)  The  subject-matter  of  this  article  is  to  perform  the 
reverse  operations  of  Art.  45 ;  i.  e.  To  find  what  part  one  quantity 
is  of  another. 

EXAMPLES. 

1.  What  part  of  a  pound  is  3  shiUings  and  4  pence  ? 

3  s.  4  d.=  40  pence; 
£1=20  s.  =240  pence. 

Now  40  pence  is  obviously  _4j)_  of  a  £,  or  1  of  a  £. 

What  part  of  a  yard  is  3  inches'!  Manifestly  JL=J_.  From 
the  above  we  draw  the  following 

Rule.  Reduce  the  given  quantity  to  the  lowest  denomination 
mentioned^  for  the  numerator ,-  then  reduce  the  integer  to  the  same 
denomination,  for  the  denominator  of  the  required  fraction. 

2.  What  part  of  a  yard  is  3  quarters  2^  nails'?  3  quarters  2^  nails= 
29  half  nails;  1  yard=32  half  nails.  Ans.   |9. 

3.  What  part  of  1  pound  Troy  is  7  ounces  4  penny-weights  ? 

Ans.  3 . 

4.  Reduce  3  quarters  7  pounds,  to  the  fraction  of  1  hundred  weight. 

Ans.  ^«^.,.. 

5.  What  part  of  2  rods  is  4  yards  1^  feet?  Ans.  ^^. 


90  ARITHMETIC. 


6.  What  part  of  1  bushel  is  l!l  pecks?  Anf:.  %. 

7.  Reduce  l:J  hundred  weight  3  quarters  20  pounds  to  the  fraction 
of  a  ton.  Am.  12. 

56 

When  the  proposed  quantity  is  not  to  be  compared  to,  or  measured 
by,  an  integer,  ike  same  principle  is  observed.  The  following  exam- 
ples will  illustrate : 

8.  What  part  of  4  hundred  weight  1  quarter  24  pounds,  is  3  hun- 
dred weight  3  quarters  17  pounds  8  ounces  ?  Ans.  g. 

9.  A  man  bought  a  piece  of  cloth  containing  8  yards  3  quarters; 
it  took  2  yards  2  quarters  to  make  a  coat ;  what  part  of  the  whole 
piece  did  it  take  1  Ans.  ^. 

10.  A  man  has  a  journey  to  perform  of  35  miles  7  furlongs ;  after 

having  traveled  15  miles  3  furlongs,  what  part  of  the  journey  has  he 

performed?  Ans.  ill. 

*  2  8  i  • 

(AuT.  47.)  To  reduce  fractions  from  a  lower  denomination  to  an 
equivalent  fraction  of  a  higher  denomination,  we  need  only  to  observe 
the  following  process : 

1 .  V/hit  part  of  a  foot  is  h  of  an  inch  ? 

Now,  to  reduce  inches  to  feet,  we  have  only  to  divide  by  12,  and 
this  we  mu!-t  do,  whatever  be  the  number  of  inches,  or  parts  of  an 
inch.     Hence,  we  must  divide  5  by  12,  which,  by  Article  44,  must  be 

J_,  Ans. 

3  6 

2.  Reduce  A  of  a  cent  to  the  fraction  of  a  dollar.  To  reduce  cents 
to  dollars,  we  must  divide  by  100.     Hence,  _  j._  =  _l_  of  a  dollar, 

•'  6  50120  ' 

the  answer. 

The  pupil  will  observe  from  this,  that  this  fraction  must  be  dvided 
at  once,  or  in  succession,  by  the  number  of  units  between  the  higher 
and  lower  denomination  mentioned. 

In  the  following  operations,  considerable  canceling  can  be  used. 

EXAMPLES. 

1.  Reduce  j  of  an  ounce  Troy,  to  the  fraction  of  a  pound. 

Ans.  J_. 
3  6 

2.  Reduce  '  •  0   of  a  minute  to  the  fraction  of  a  day. 

Ans.  _^_. 


3.   Reduce    ^-i_   of  a  pound  avoirdupois,  to   the  fraction  of  a 

ns. ?_^, 

1. 

Ans.  _3_ 


hundred  weight.  Ans. ?_^. 

4.  Reduce  f  of  a  nail  to  the  fraction  of  an  ell  English. 


5.  Reduce  _V.  of  a  penny  to  the  fraction  of  a  pound. 


Ans.  ^--. 
576  0 


VULGAR    FRACTIONS.  91 


6.  Reduce   i    of   a    pint    to   the  fraction  of  a  bushel. 


Ans.  -i-. 

3   2  0 


7.  Reduce  .?_  of  an  ounce  to  the  fraction  of  a  hundred  weight. 


A)is. 

9  8  se 


8.  Reduce  1  of  an  inch  to  the  fraction  of  an  ell  English. 

Ans.  _L. 

7  5 

(Art.  48.)  To  reduce  a  fraction  from  one  denomination  to  an- 
other, on  the  same  scale, 

Reduce  7  to  fourths.  We  write  I,  and  then  multiply  numerator 
and  denominator  by  4,  making  2_8j  ^^'hich  is  still  7  in  value.  (See 
Art.  39.) 

In  the  same  way,  let  us  reduce  4  to  fourteenths ;  or,  in  other 
words,  change  |.  to  fourteenths^. 

5 
6" 

We  write  '{  and   multiply  numerator  and   denominator    by    14 

14  14         14         14 

2.  How  many  fifths  are  (here  in  i  of  anything?  Ans.  4|. 

3.  How  many  _L   are  there  in  J-  of  any  number,  or  any  thing? 

'"*  Ans.    6f- 

This  is,  anything  divided  into  24  equal  parts,  6|  such  parts  is  the 
same  portion  of  the  whole  ais  -|-''j.is  of  1. 

4.  Reduce  1  to  a  fraction  whose  denominator  shall  be  4. 

3^ 
Ans.  p' 

4 

5.  Reduce  11  to  a  fraction  whose  denominator  shall  be  20. 

Ans.  -— f- 
20 


6.  How  many  -V  of  a  shilling  are  in  4  of  a  shilling ' 


66 

Alls.  _T 


12 
7.  How  many  pence  are  there  in  i.  of  a  shilling  1  Ans.  6£. 

N.  B.  Observe,  that  examples  6  and  7  are  substantially  the  same, 
except  that  one  calls  for  the  answer  in  parts  of  a  shilling,  the  other 
calls  for  it  in  pence. 

(AnT.  49.)     When  several  fractions  have  different  denominators. 


92  ARITHMETIC. 


it  is  important,  many  times,  to  bring  them  under  one  denomination, 
as  we  can  neither  add  nor  subtract  them  until  this  is  done.  (See 
Art.  5.)  For  example,  reduce  ^  and  §  to  a  common  denominator, 
without  charginj^  their  values.  Eighths  cannot  be  reduced  to  halves 
in  this  case,  but  halves  can  be  reduced  to  eighths,  by  multiplying  nu- 
merator and  denominator  by  4,  and  then  wc  have  ±  and  f ,  8  being 
the  common  denominator. 

Thus,  when  the  deiiominators  are  muliiples  of  each  other,  and  the 
pupil  has  a  clear  understanding  of  the  nature  of  a  fraction,  he  can 
reduce  them  all  to  one,  or  a  common  denominator,  by  multiplying  or 
dividing   numerators    and   denominators,  as   judgment  may  dictate. 

EXAMPLES. 

1.  Reduce  |,  _s_,  J'_,  to  equivalent  fractions,  having  a  common 
denominator.  Ayis.  ^s_^  _6_^  _5_, 

2.  Reduce  1  _?_,  J'-,  to  their  equivalents,  having  a  common  de- 
nominator.  "  Ans.   _2_,  _5_^  _8_. 

All  rules  for  reducing  fractions  to  a  common  denominator  must  be 
based  on  the  principle  that,  numerators  and  denominators  multiplied 
or  divided  by  tlie  same  number,  does  not  change  the  value  of  the 
fraction.  On  this  principle,  we  may  understand  the  following  rule, 
which  is  universal  in  its  application : 

Rule.  Multiply  each  numerator  irito  all  the  denominators  ex- 
cept its  0W71,  for  a  new  numerator.  Then  multiply  alt  the  deno- 
minators together  for  a  new  denominator,  and  place  it  under  each 
new  7iumerator. 

1.  Reduce  -f,  2,  and  ^,  to  a  common  denominator. 

2X2X4^=16  the  numerator  for  -f. 
1X3X4=12  -  do.  -  |. 
3X3X2=18  -  do.  -  |. 
3X2X4=24  the  common  denominator. 


The  equivalent  fractions  are 


{1  6-_2 
24— 3* 
1  2-_l 
24  2* 
2  4  4* 


It  is  evident  that  the  values  of  the  fractions  are  not  changed,  be- 
cause both  the  terms  of  each  are  multiplied  by  the  same  number. 

2.   Reduce  1  and  1  to  a  common  denominator. 

Ans.    J    63  7' 


VULGAR    FRACTIONS.  93 

(AiiT.  50.)  To  follow  the  above  rule  in  all  cases  might  often 
require  more  niechauical  labor  than  necessary ;  and  to  abridge  the 
operation  we  must  lind,  not  merely  the  common,  but  tlie  least  com- 
mon denominator. 

Preparatory  to  this,  we  must  be  able  to  find  the  least  commnn  mxil- 
tiple  of  tw!)  or  more  numbers  ,-  that  is,  the  least  number  divisible 
bij  them  icUlwut  remainders. 

RfLK.  Write  the  numbers  one  after  the  other,  and  draw  a  line 
beneath  them ;  then,  take  any  prime  nund)er  which  will  divide 
two  or  more  of  them  tvithout  remainder,  and  divide  all  the  num- 
bers tiiat  will  so  divide  —  ivriting  the  quotients  beneath,  and  all  the 
numbers  that  are  not  divisible  by  it.  Find  a  prime  number  that 
will  divide  two  or  more  numbers  in  this  second  line,  and  proceed  as 
bpfore.  Continue  the  operation  tmtil  there  are  no  two  numbers  left 
bavins;  a  common  divisor;  then,  multiply  all  the  divisors  and  re- 
maining  numhers  together,  and  their  product  will  be  the  least  com- 
mon multiple  sought. 


1.   Let  it  be  required  to  find  the  least  common  multiple  of  12,  15, 
7,  18,  3,  5,  and  35. 

7 

5 

3 

2 


7X5X3X3X2X2=1260. 

First,  dividing  by  the  prime  number  7,  we  find  that  it  divides  two 
of  the  given  numbers,  7  and  35  —  the  rest  being  written  in  the  line 
below.  We  take  5  for  the  next  divisor,  which  di'^des  15  and  5. 
Continuing  the  operation  in  the  same  manner,  we  divide  successively 
by  3  and  2,  when  the  division  can  be  carried  no  further.  Then,  mul- 
tiplying all  the  divisors  together,  and  that  product  by  the  undivided 
prime  numbers  left,  and  the  result  in  this  case  is  1260.  It  is  evident 
that  this  must  be  the  least  common  multiple,  because  it  contains  07ily 
the  prime  factors  necessary  to  make  up  the  given  numbers.  The 
same  must  be  true  of  any  other  numbers. 

2.  Find  the  least  number  that  is  divisible  by  9,  12,  16,  24,  36, 
without  remainders.  Ans.  144. 


12, 

15, 

7, 

18, 

3, 

5, 

35, 

12, 

15, 

1, 

18, 

3, 

5, 

5, 

12, 

3, 

1, 

18, 

3, 

1, 

1, 

4, 

1, 

I, 

6, 

1 

1, 

1, 

2, 

1, 

1, 

3, 

1, 

1, 

1. 

94  ARlTHMliTlC. 


N.  B,  In  the  opeiHtion,  9  anri  12  may  I'e  leO  f^nt,  as  any  num- 
ber divislibe  by  36  will  be  divisible  by  9  and  12, 

3.  Find  the  least  number  divisible    by  each    of    the  nine  digits 

A71S.  2520. 

4.  Find  the  least  number  divisible  by  75,  50,  15,  20,  30,  and  45. 

Ans.  900. 

(Aut.  51.)  We  are  now  prepared  to  give  the  following  rule  for 
finding  the  least  common  denominator  to  any  set  or  group  of 
fractions. 

Rule.  Find  the  least  common  multiple  of  all  the  denominators. 
This  will  he  the  common  denominator.  For  the  numerators,  divide 
the  common  denominator  by  each  particular  denominator,  and  mul- 
tiply the  quotient  by  its  numerator. 

EXAMPLES. 

1.  Reduce  ^,  4,  y-,  and  ^-^  to  the  least  common  denominator. 
7 
2 


2, 

7, 

16, 

21, 

2, 

1, 

16, 

3. 

1, 

1, 

8, 

3. 

7X2X8X3=336,  least  common  denominator,  or  denominator  to 
all  the  new  fractions.  To  obtain  the  new  numerators,  divide  this  by 
the  former  dtnominators,  and  multiply  the  quotient  by  the  former 
numerators,  thus: 

336X1 

— =168= 1st  numerator. 

2^L>lf=192=2d  numerator. 
7 

336X3       „ 

— — — =63=3d  numerator. 
16 

336X3 

— _  - — =32=4th  numerator. 

The  new  fractions  equivalent  to  the  given  fractions  are  l^lt  i-^, 

J5  3_     Ji2_ 
3  36»    3  3  6 

2.  Reduce  _"_.  1.  1.  and  i,  to  their  least  common  denominator. 

An^.    AHl      231      3.12       1  5   1 
616'    616'    6Ifi»    6  Tfi' 

3.  Reduce  _2_,  ^^^  :|7^  and  J'-^  to  their  least  common  denomi- 
nator. "  Ans.  J^_    JJ_    Lll       8    . 

150'     I50»     150'     150 


VULGAR    FRACTIONS.  UT) 


^4.   Keduce  -^  .»_^  ^,  and  6,  to  their  least  coihuidi)  denoiniriator. 

Ans.  -s_fi_,  JJ,  1-L'l.  LJJ?. 

•J32'252'25i;»      252 

5.  Reduce  S^,  2^,  and  If,  to  their  least  common  denominator. 

AflS.    4J       l_8      IJ. 
A  ^,  8    »      8    '      S 

4         2-^ 

6.  Reduce  ^  of  ^,  7  andji  to  their  least  common  denominator. 

s  ^ 

AnS.  -8^      4  80       4  5  . 
2  4'      2  4   »    2  4 

7.  Reduce  --  of  -^, -^  of  -^  and  7,  to  their  least  common  deno- 

minator.  Ans.^-?^J    111    2_2_o_5 

3Ts  '  3  1  5'    315' 

ADDITION  OF  FRACTIONS. 

(Art.  52.)  We  cannot  add  unlike  things  together,  (Art.  .5,) 
therefore,  betbre  fractions  can  be  add(ul,  we  must  be  positive  that  they 
not  only  refer  to  the  same  integer,  (lielong  to  the  same  scale  of  mea- 
sure.) but  have  the  same  denomination  of  that  scale.  For  illustra- 
tion, we  cannot  immedintely  add  together  f  of  a  foot  and  -J  of  an 
inch,  for  tlie  s.uiie  reason  that  we  cannot  put  feet  and  inches  into  the 
same  .sum.  Equally  impossible  is  it  to  add  fourths  and  thirds  toge- 
ther. Hence,  all  the  difiiculty  a  pupil  will  meet  with  will  be  in  mak- 
ing the  preparations  to  add,  and  not  in  adding. 

When  all  the  preparations  are  made,  the  fractions  are  added  by 
adding  together  the  numerators,  and  setting  the  common  denominator 
under  the  sum  ;   for,  the  sum  of  ^,  £,  and  g,  is  ccrtaiiily  il. 


Take  the  second  example  of  the  last  article  (51),  and  require  the 
in  of  _?_^  |,  ^,  and  i- 

The  equivalent  fractions  having  a  common  denominator,  we  see  by 
"erring  back  to  article  51,  are  A?l.  'l'±l.,  252     \_b_i^, 

°  (>16'016'616»G16 

Their  sum  is  504-|-23I-|-252-4- 154=1 141. 

Am.  »-ij'_i 

6  16' 

From  the  foregoing  we  deduce  the  following  general 


Rule.  Reduce  ihe  fractions,  if  necessary,  to  the  same  standard 
or  integer  of  measure  ,-  compound  fractions  to  single  ones,  and  ail 
to  a  common  denominator,-  then,  add  the  numerators,  and  place 
their  sum  over  the  common  denominator. 

When  several  fractions  have  denominators  that  are  multiples  of 
each  other,  as  i,  i,  |,  &c.,  we  can  reduce  them  to  a  common  denomi- 
nator without  any  formality  of  rule,  by  simply  multiplying  the  nume- 
rators and  denominators  of  those  fractions  having  the  smaller  denomi- 
nators, by  such  numbers  as  will  bring  the  denominators  alike.  'J'hus, 
in  the  above,  ^=i>  4=|,  and  |  stands:  and  their  sum  is  -1=1^. 


96  ~  ARITHMETIC. 


In  the  same  manner  we  find  the  sum  of  the  fractions  ^1,  |,  | 


and  ^1,  to  be  Il^Ve* 

A  similar  operation  will  give  the  sum  of  the  following  fractions : 
L-J-l-Ul-i--'^—     It  is  evident  that  these  denominators  measure  24; 

2  T^  6    1^  n  n^  1  2 

that  is,  all  of  them  will  divide  24 ;  hence,  we  can  change  all  the  frac- 
tions to  twenty-fourths,  by  mere  inspection.     Thus,  1  =  12    i  =_4 

3  =  _9_    _"_  =  !,£.      Sum,  3  9._1|. 

8  24'     12  24  24  " 

When  we  have  mixed  numbers,  add  the  whole  numbers  separately, 
and  then  add  the  fractions.     The  two  sums  put  together  will  be  the 

sum  required. 

EXAMPLE. 

Add  3i,  Ih  21,  ry^,  9tV,  together.  Integers,  3-l-7-}-2-i-74-9 
=29.  Fractions,  « |+_n_-l-_«_-l--3_+/-=||=|.  '  Whole 
sum,  298. 

EXAMPLES    FOR    PRACTICE. 

1.  Add  I  and  2.  Ans.  3  5. 

2.  Add  I,  I ^  and  5^  together.  .4ns.  li.||. 

3.  Add  I,  i,  2^  and  ^  of  |,  together.  Ans.  \^^-. 

4.  Add  together  ^^^^J^-^-  Ans.  U t  p. 

5.  Add  I  of  i  of  |,  to  6|  of  i,  and  f  of  je ,  together. 

ylns.  4JL. 

I  8 

6.  Add  I  of  a  rod  to  |  of  a  foot.  Ans.  13^  feet. 

7.  Add  i  of  a  mile,  ■§  of  a  furlong,  and  |-  of  a  rod  together. 

Ans.  307|.  rods. 

8.  Add  ^-  of  3^,  _?_.  of  2^,  -?_  of  U,  _i_.  of  2|,  and  !)_  of  3^, 
together.  ^W5.  I. 

[This  last  example  is  very  simple  and  brief.] 

9.  What  is  the  sum  of  ^  of  a  pound  and  |^  of  a  shilling  ? 

^7W. 'i|5=13s.  lOd.  2|q. 

10.  What  is  the  sum  of  §  of  a  yard  and  ■§  of  a  quarter  ? 

Ans.  I_i_  yards. 

11.  What  is  the  sum  of  2.  of  a  ton,  added  to  ^   of  a  hundred 

3  / 

weight?  A'^is.  123  cwt. 


SUBTRACTION    OF    FRACTIONS.  9' 


12.  What  is  the  sum  of  g  of  a  day  added  to  ^  of  an  hour  ! 

Ana.  9t,  hours. 

13.  What  is  the  sum  of  f  oi  a  pound  and  ^  of  a  shilling] 

Ans.  3s.  2d. 

14.  What  is  the  sum  of  1  of  a  week,  |  of  a  day,  and  i  of  an 
hour?  Ans.  1  d.  22  h.  15  m. 


SUBTRACTION  OF  VULGAR  FRACTIONS. 

(Art.  53.)  We  would  remind  the  pupil  that,  in  addition,  we  took 
the  sum  of  the  numerators,  after  the  fractions  were  reduced  to  a  com- 
mon denominator.  Hence,  the  difference  of  the  two  fractions  must 
1)3  found  by  taking  the  difference  of  their  numerators,  when  the  deno- 
minators are  ahke.  For  example,  the  difference  between  _"_  and  _?_, 
must  be  -?  _  i  and  the  difference  between  f  and  f  niusl  be  f,  6lc. 
These  observations  must  give  us  the  followmg 

RuLK.  Prepare  ihe  fractions  as  for  addition,  and  place  the  dif- 
ference of  the  numerators  over  ihe  common  denominator,  fir  the  dif- 
ference tf  the  fractions. 

EXAMPLES. 

1.  Find  the  difference  of  t  and  J- .  Ans.    s 

0  12  I  -i 

2.  From  #  take  1.  Ans.  _«_. 

*  7  £  8 

3.  From  6|  yards  take  3|  yards. 

Rules  would  reduce  these  numbers  to  improper  fractions,  and  then 
to  a  conjmon  denominator,  thus:  ^J  and  3J*,  difference  -J=z'2,^. 
But,  a  little  tact,  which  should  always  be  cultivated,  will  do  it  more 
elegantly,  thus:  6|=5'Jj  and  34=31;  hence,  2^  is  their  dilier- 
ence.      In  like  manner  perform  the  two  following  : 

4.  From  9L  take  Q^_.  Ans.  2J_. 

5  10  10 

5.  From  I9i  vards  take  12_?_  yards.  Am.  6L?.. 

6.  What  is  the  difference  between  f  of  b\  and  |  of  4i. 

Ans.  1-2. 

■) 

7.  What  is  the  difference  betv/een  ^  of  a  rod  and  |  of  a  foot  ? 

Ans.  1^  feet. 

8.  From  ^  of  a  yard  take  f  of  a  quarter.  Ans.  ^  quarter. 

9.  From  i.  of  a  pound  take  -f  of  ^  of  a  shilling. 

"  Ans.  £i_i=10s.  7d.  Uq. 

3  6  0  ^  ^ 

10.  From  f  of  a  yard  take  J  of  a  quarter. 

Ans.  'J'=l  qr.  3  na. 


98  ARITHMETIC. 


11.  From    7  ells   English    take    4J    yards. 

Ans.  32  ells  English. 

12.  From  1  of  a  pound  sterling  take  |  of  a  penny, 

/  *  Ans.  2s.  iLld. 

13.  From  1  of  a  rod  take  -f  of  a  foot.         Aiis.  3J    nr  10^  feet. 

14.  From  1  of  an  ounce  take  -|  of  a  pennyweight. 

A7if!.  ''_5j»  dwt.,  or  16LL  dwt. 

2  3  '  2  S 


MULTIPLICATION  OF  FRACTIONS. 

(Aht.  54.)  Multiplication  is  the  same  in  its  nature,  whatever  may 
be  the  value  of  the  numbers  multiplied  together. 

When  we  consider  two  terms  or  factors  to  be  multiplied  together, 
a  multiplicand  and  a  mulliplier,  the  multiplicand  must  be  taken  as 
many  times  as  there  are  units  in  the  multiplier.  When  there  is  not 
a  unit  in  the  multiplier,  then  we  are  to  multiply  by  a  fraction,  and  it 
is  taking  the  multij)licand  part  of  one  time,  and,  of  course,  the  product 
will  be  less  than  the  multiplicand. 

For  example :  Let  it  be  required  to  multiply  2.  by  2.  Here  1 
is  to  be  considered  the  multiplier,  which  is  twice  i.  Now,  we  must, 
obviously,  take  ^^  4,  of  1  time,  —  that  is,  divide  it  by  7 ;  which  is,  by 
Article  44,  J-.  But,  we  want  |,  twice  as  much  as  1  :  hence,  JL. 
must  be  the  required  product.  And  this  we  should  have  obtained  by 
the  following 

Rule.  To  multiply  one  fraction  by  another.  Multiply  the  nu- 
merators together  for  the  numerator  of  the  required  fraction,  and 
the  denominators  together  for  the  denominator  ;  always  canceling, 
when  possible. 

N.  B.  Mixed  numbers  must  be py-eviously  reduced  to  improper 
fractions. 

Another  explanation  : 

1.  Multiply  I  by  |.  Ans.  /^==y\. 

3X2     6 
Here  '-—  = — •     But,  the  multiplier  is  only  the  l  part  of  2  ;  con- 
4        4  * 

sequently,  i  must  be  diminished  corre-spondently,  by  multiplying  its 

denominator  by  5. 

2.  Multiply  2^  by  I  of  I  of  8. 

5X3X2_X8^X2_e^^ 
2X4X5X1         1 


MULTIPLICATION    OF    FUACTIONS.  09 


An.s.  4^. 

f    by   7i    by    ^^„ 

by    7. 

Atis.  23. 

,  and  4_''_. 

^n.5.  2J_. 

J  0 

1.  Multiply  5  by  -f  of  ^.  ./1/J5.  1. 

2.  Multiply    J*-,   by  ^,   and   that   product   by  7^-. 

3.  Required    the  product   of    21    by 

4.  Required  the  product  of  1 

5.  Multiply  together  5,  f,  1  of  li,  and  41.  yl/?.s.  sj'  =71. 
G.   Multiply  fi   of  3  times  _9_  of  5  times  ^  of  ^  of  i,  by  1  of  i 

of   A   of   J    .  7lrt5.  g^. 

7.  What  will  8^  pounds  of  tea  cost  as  1^^  dollars  per  pound  1 

A71S.  1011  dollars. 
1  6 

8.  What  will  4|  cords  of  wood  cost,  at  3|  dollars  per  cord  ? 

Ans.  1713  dollar?. 

1  6 

9.  If  a  horseman  travel  at  the  rate  of  7^  miles  an  hour,  for  6_"_ 
hours,  how  far  will  he  have  gone?  Ans.  51 -U  miles. 

10.  What  is  the  cost  of  9^  yards  of  linen,  at  $lf  per  yard  ? 

Ans.  $12.15. 

^        6  4 


DIVISION  OF  FRACTIONS. 

(AuT.  .5.5.)  The  essential  nature  of  division  is  the  same,  whether 
the  dividend  and  divisor  are  whole  numbers  or  fractions.  It  is,  sim- 
ply, finding  how  many  times,  or  what  part  of  a  time,  one  num- 
ber is  contained  in  another. 

For  example:  We  may  demand  that  6  shall  be  divided  by  j>,  and 
an  inattentive  pupil  might  suppose  that  we  required  tlie  half  of  6  ; 
but  this  is  not  true.  We  demand  how  many  ti'iies  ^  is  contained  in 
6:  or,  we  might  say.  Divide  6  into  equal  parts,  and  have  each  part 
equal  to  {j ;  how  many  parts  will  there  be  ]  Ans.  12.  Divide  6  by 
^.   Am.  18.     Divide  6  by  J_.  Ans.  60.     Divide  6  by  |.  Ans.  42. 

Divide  6  by  2.  Ans.    As  1  is  contained  in  6  42  times,  2  will  be  con- 

•'7  7  7 

tained  21  times.     That  is.  We  multiply  the  dividend  by  the  denomi- 
nator of  the  divisor,  and  divide  the  product  by  the  numerator. 

When  the  dividend  is  a  fraction,  the  result  may  be  obtained  by  the 
following 

Rule.  To  divide  by  a  fraction,  invert  the  divisor,  and  proceed 
as  in  muUiplicutidn. 

N.  B.  When  the  divisor  is  a  compound  fraction^  invert  every 
ierrii  ,•  and,  in  all  cases,  caticel  as  much  as  possible. 


100  ARITHMETIC. 


EXAMPLES. 

3 

21,3 
Form,  — ^— 

Canceled 

#,3 

9. 

^??s. 

48. 

u47lS. 

117. 

-4/25 

49. 

^ns 

.99. 

I.         Form  -^ 
5                       4. 

4. 

4. 

5. 

'3: 

1.  Divide  21  by  |. 

2.  Divide  18  by  f. 

3.  Divide  63  by  _7_. 

4.  Divide  42  by  ^. 

5.  Divide  66  by  |. 

6.  Divide  f  of  ±  by  i  of  I 

Canceled  ■     ^'  ^'        —  ;  therefore,  4  is  the  answer  required. 
^.  A-  1-  ^• 

7.  Divide  4  of  _«_  of  ^,  by  i  of  6  of  -"_.  yl"5.  4. 

8.  Divide  ]^^  of  I  of  1.^  of  1-f  of  |8  of  i,  by  ^  of  i  of  _P_  of  4- 

Ans.  m=l2J-. 
4  1  3  4  7  o 

9.  Divide  4^  by  i  of  64.  Ans.  _'/-,.. 
10.  Divide  6|  by  |  of  42.  A«5.  -jP_. 

.11.  Divide  f  of  I  by  4  of  6^.  Ans.  ^^-. 

(Art.  56.)  Fractions  must  refer  to  the  same  scale  of  weight  or 
measure,  before  we  can  divide  one  by  the  other.  For  instance,  we 
cannot  divide  the  fraction  of  a  gallon  by  the  fraction  of  a  pint,  or  by 
the  fraction  of  a  gill,  before  we  reduce  one  to  the  denomination  of  the 
other,  for  the  same  reason  that  we  cannot  add  and  subtract  unlike 
quantities,  (Art.  5.) 

EXAMPLES. 

1.  How  many  times  is  •§  of  a  pint  contained  in  1  of  a  gallon  1 

Ans.  6^. 

2.  How  many  times  can  a  vessel  holding  _P_.  of  a  quart,  be  tilled 
from  5  of  a  barrel,  containing  31^  gallons  ?  Ans.  4G|. 

3.  How  many  rods  are  there  in  60^  yards  ?  Ans.  1 1. 
*  4.  How  many  times  is  |  of  an  inch  contained  in  1  of  a  yard  ? 

Ans.  3i2. 
5.  How  many  times  is  1  of  a  rod  contained  in  ^  of  a  foot  ? 

A71S.    J-. 

»  9 


PRACTICAL    APPLICATION    OF    FRACTIONS.  101 


PRACTICAL  APPLICATION  OF  FRACTIOxNS. 

1.  What  will  51  yards  of  calico  cost,  at  31^  cents  per  yard  ? 

Alt.'.  '^pl.CA-{- 

2.  What  will  S^  yards  of  linen  cost,  at  G2^  cents  per  yard  ? 

3.  What  will  5-}  yards  of  linen  cost,  at  Sl:^-  cents  per  yard  ? 

An.v.  S4.67-I- 

4.  Bought  a  yard   of  cloth   for  ^  of  a  dollar ;  what  would  be  the 
cost  of  8  yards,  at  the  same  rate  ?  Ans.  $6^. 

5.  If  I  pay  I  of  a  dollar  for  a  gallon  of  oil,  what  will  I85  gallons 
cost?  '  Ans.^i^^. 

6.  What  would  Jl  of  an  acre  of  land  cost,  if  ^  of  an  acre  cost  5^27. 

Ans.  $48. 

7.  Bought  f  of  a  fifteen-acre  lot,  and  sold    |  of    what     I     pur- 
chased; how  much  did  I  sell?  Ans.  4^  acres. 

8.  What    would   395   pounds  of  sugar  cost,  at  ?  of  ^  of  a  dollar 
per  pound?  "       Ans.  $52^. 

9.  Add   i    of   a   pound  sterling  to   ^   of    ^    of    a  shilling. 

Ans.  2s.    10|d. 

10.  Add  I  of  a  mile,  ^  of  a  furlong,  and  J5_  of  a  rod  together. 

Ans.  233^  roJs. 
IL  Add   1    of  a   hogshead   to   ^    of   a   gallon. 

Ans.  54-^-  gallons. 

12.  What  is  the  value  of  »  of  a  day?  Aiis.  10  h.  17^  m. 

13.  What  is  the  value  of  1  of  a  dollar?  Ans.  421  cts. 

14.  Reduce  |  of  |.  of  a  pound,  avoirdupois,  to  integers. 

Ans.  4  oz.  11|3  dr. 

15.  From  J-  of  a  ton  take  f  of  a  hundred  weight. 

A71S.  -\  cwt. 

16.  If  ^  of  a  dollar  pay  for  3|  pounds  of  sugar,  how  many  cents 
is  it  per  pound  ?  Ans.  10. 

17.  If  $20J  pay  for   16^  barrels  of  apples,  what  is  the  cost  per 
barrel  ?  Ans.  $1 ,26,  nearly. 

18.  If  ^  of  2  of  a  farm  cost  1200  dollars,  what  would  the  whole 
cost  1  Ans.  §7200. 

19.  If  6  pounds  of  tea  cost  9^  dollars,  what  is  it  per  pound  ? 

Ans.  Sl|. 

20.  If  5^  yards  of  cotton  cloth  cost  §  of  a  dollar,  how  many  cents 
is  it  per  yard  ?  Ans.   15_5_. 

21.  If  ^  of  a  dollar  pay  for  1|  yards  of  cloth,  what   is  the  price 
per  ell  English.  Ans.  79_.fi_. 

22.  If  a  piece  of  cloth  measuring  13f  yards,  cost  27^  dollars,  what 
is  the  value  of  1  yard  1  Ans.  |2_2_. 


102  arithmp:tic. 


23.  If  7  horses  consume  2|  tons  of  hay,  how  much  does  each  con- 
sume ?  Ans.  11. 

24.  What  part  of  a  rod  is  2^  yards  1  A?is.  _5_. 

25.  Find  the  difference  between  3  of  an  ell  English,  and  t  of  3 
feet?  Ans.  IJ_. 

5  6 

Questions. 

How  do  you  judge  of  the  value  of  a  fraction  ? 

What  numbers  do  you  compare  in  making  up  your  judgment? 

]f  you  double  a  denominator,  what  effect  does  it  have  upon  the 
fraction  ? 

If  you  double  a  numerator,  what  effect  does  it  have? 

If  you  double  both  numerator  and  denominator,  what  effect  ? 

If  you  add  the  same  number  to  both  numerator  and  denominator  of 
a  fraction,  does  it  change  its  value? 

How  do  you  reduce  complex  fractions  to  simple  ones  ] 


DECIMAL  FRACTIONSr 

(Art.  57.)  The  common  or  vulgar  division  of  things  into  halves, 
quarters,  thirds,  &c.,  &c.,  so  convenient  and  proper  in  the  every-day 
little  concerns  of  business  life,  becomes  troublesome  and  perplexing 
when  we  have  a  variety  of  these  different  denominations  to  add  or 
subtract  in  large  mathematical  calculations;  and,  to  surmount  this 
difficulty,  we  have  a  more  artificial  division  of  things  into  tenths,  those 
parts  again  into  tenths,  and  again,  if  need  be,  these  svibdivisions  into 
tenths,  &c.,  Sec. ;  thus  having  a  decreasing  order  of  fractions  on  the 
same  graded  scale  as  natural  numbers,  ten  of  one.order  making  one 
of  another  ;  and,  of  course,  we  may  call  them  decimals',  from  deci- 
mils,  the  Latin  word  meaning  tenths.  From  this  it  is  evident  tliat,  if 
we  divide  a  tenth  into  ten  equal  subdivisions,  one  of  them  will  be  one- 
hundredth  of  the  unit,  &c.,  &c. ;  hence,  decimal  fractions  can  have 
no  other  denominators  than  10,  100,  1000,  &c.,  &c.  This  being 
systematized  and  properly  understood,  the  denominators  need  not  be 
written.  If  the  fraction  is  tenths,  one  place  from  the  unit  wdl  ex- 
press it,  with  a  point  between  ;  if  hundreds,  or  two  orders  from  the 
unit,  two  places  of  figures  must  be  taken  :  thus,  _i_  is  expressed  .05, 
_^_-_  thus,  .005,  always  as  majiy  places  as  the  understood  denomi- 
naior  contains  ciphers. 

Thus,  7_s'_  is  written  in  decimals, ....  7.5. 

4tVo' 4.24. 

I^Vt/WV' 141-1356. 


DECIMAL    FRACTIONS.  108 


Ciphers  written  niter  a  decimal  neither  increase  nor  decrease  it,  for 
this  does  not  change  the  place  of  the  figures.  'I'hus,  2^'^  is  2.6,  and 
2.60000  is  the  same  value  —  the  6  still  being  in  the  same  orc/e?- of 
numbers;  but,  ciphers  on  the  other  side,  as  2.06,  changes  the  orc?er, 
and  diminishes  the  value  ten-fold  for  every  remove. 

The  following  shows  the  scale  : 


■±3         «       -r: 


P 


J3  '^ 


C 


llllllllJllia 

333333   3,   333333 

Express  the  following  in  figures. • 

1.  Sixty -six  and  seven-tenths.  Ans.  66.7. 

2.  One  and  thirty-five-hundredths.  Ank:  1.35. 

3.  Eighty-one  and  1  ten-thousandth.  ^ws.  81.0001. 

4.  One  hundred  and  67  ten-thousandths.  Ans.  100  0067. 

5.  Three  and  three-hundredihs.  Ans.  .S.03. 

6.  Seventy-five  and  75  ten-thousandths.  Atis.  7f».C075. 
Let  it  be  observed,  that  these  fractions  vary  in  value  on  the  same 

scale  as  whole  numbers  ;  and  the  denominations  of  Federal  Money 
are  nothing  more  than  whole  numbers  and  decimals,  the  (lollars  being 
units,  and  the  cents,  mills,  &c.,  being  decimals  of  a  dollar,  as  men- 
tioned in  Article  8. 

To  familiarize  decimals  to  Federal  Money,  let  the  pupil  make  the 
following  reductions  : 

1.  Reduce  §41.7  dimes  and  6  mills,  to  mills.  ^1;?.*.  4I70G. 

2.  Reduce  31806   mills    to   dollars,   cents,   and    mills. 

Am.  31.806. 

3.  Reduce  $10,  3  dimes  4  cents  and  9  mills  to  mills. 

A)is.  10349. 

4.  Reduce   12349  mills  to  dollars,  cents  and  mills. 

J»5.  $12,349. 

5.  Reduce  $25  to  cents  Ans.  2500. 


ADDITION  OF  DECIMALS. 

(Art.  58.)     We  again  call  to  mind  the  fact  mentioned  in  Article 
5,  that  only  like  things  can  be  added  together.     In  whole  numbers, 


104  ARITHMETIC. 

units  can  only  be  added  to  units,  tens  to  tens,  &c.  So  in  decimals  : 
tenths  can  only  be  added  to  tenths,  &c.  Hence,  we  have  the  follow- 
ing rule  for  addition  in  decimals: 

Rule.  Set  down  the  numbers  to  be  added  so  that  tenths  shall 
fall  under  tenths,  hundredths  under  hundredths,  6(c.  Tliis  will 
brins;  all  the  decimal  points  directly  under  each  other ,-  add  up  as  in 
wh(jle  numbers,  and  place  a  separatrix  between  the  units  and 
decimal. 

We  shall  give  but  very  few  examples  under  this  rule,  as  the  opera- 
tion, (except  the  case  in  placing  the  decimal  points  under  each  other,) 
is  the  same  as  in  addition  of  whole  numbers. 


EXAMPLES. 

Add  37.03,  621.57,  .521,  20.007. 

(2) 

Operation.  Operation. 

37.03  .199 

621.57  2.7569 

.521  ,25 

20.007  .654 


Sum,  679.128  Sum,  3.8599 


3.  What  is  the  sum  of  two-tenths,  two-hundredths,  and  two-thou- 
sandlhs  ?  Ans.  .222. 

4.  What  is  the  sum  of  14,  and  16  ten-lhousandths? 

Ans.  14.0016. 

5.  What  is  the  sum  of  twenty-six  and  twenty-six  hundredths,  sev- 
en-tenths, and  six  and  seventy-three  hundredths,  four  and  four  thou- 
sandths ]  Ans.  37.694. 

6.  What  is  the  sum  of  9  dollars  3  mills,  14  dollars  3  dimes  9  cents 
1  mill,  104  dollars  9  dimes  9  cents  9  mills,  999  dollars  9  dimes  1  mill, 
4  mills,  6  mills,  and  I  mill  1  ylns.  §1 1"28..^05. 

7.  What  is  the  sum  of  4  dollars  21  cents,  87^  cents,  3  dollars  6^ 
cents,  2  dollars  2  cents,  5  dollars  372  cents,  and  37  dollars  7  cents  1 

Ans.  $52.16i. 


SUBTRACTION  OF  DECIMALS. 

(  Aut.  59.)  From  the  very  nature  of  these  numbers,  as  previously 
explained,  we  must  have  the  following 

Rule.  Place  the  numbers  as  in  addition  of  decimals,  subtract  as 
in  whole  numbers,  and  place  the  decimal  point  in  the  remainder, 
under  the  decimal  points  above. 


DECIMAL    FRACTIONS.  105 


EXAMPLES. 

1.  From  91.73  take  2.138. 

91.73  (2)   2.73 

2,133  1.9185 


/l/i5.  89.592  Ajis.     .8115 

3.  Find  the  (Utrerence  between  714  and  .916.      Ans.   713.0S4. 

4.  From  .145  take  .09GS4.  Ans.  .04816. 

5.  How    much    greater  is    2    than    ,298?  Ans.  1.702. 

6.  What  is  the  ditierence  between  75  dollars  and  75  cents  ? 

7.  From  21.004  take  75  hundredths.  Ans.  20.254. 

8.  From  260.4709  take  154.0578.  yln.s6,4131. 

9.  From  10.0302  take  2  ten-thousandths.  Ans.  10.03. 
10.  From  2.01  take  99  hundredths.  Ans.  1.02. 

MULTIPLICATION  OF  DECIMALS. 

(.Art.  60.)  Vie  propose  now  to  investigate  a  rule  for  the  multi- 
plication of  decimals;  and  as  these  numbers  vary  as  simple  num- 
bers, ten  of  one  order  making  one  of  a  superior  order,  the  actual  mul- 
tiplication of  the  figures  must  be  the  same  as  taught  under  simple 
multiplication  :   it  only  remains  to  point  out  the  value  of  the  product. 

Let  us  multiply  .3  by  .5.  To  understand  the  operation,  make  vul- 
gar fractions  of  them,  and  we  have  J^-  to  multiply  by  ^-.  The 
product  (Art,  54)  is  JJL  ;  or,  by  the  definition  of  decimals,  .15. 

Again:  Multiply  .oVby  .13,  or  _^^_  by  _y_.  Product,  __s_2__, 
or,  by  the  definition  of  decimals,  .0052. 

Psow,  we  perceive  that  every  place  in  a  decimal  has  a  cipher  cor- 
responding in  a  denominator  understood ;  and  the  multiplication  of 
these  denominators  make  as  many  [)laces  as  there  are  ciphers  (Art. 
15)  ;  that  is,  as  many  as  there  are  places  of  decimals  in  both  the  fac- 
tors.    Hence,  we  have  the  following 

RuLF..  Multiply  as  in  simple  niimhers,  and  point  off  in  the  pro- 
duct, from  the  right  hand,  as  many  figures  fur  decimals  as  there 
are  decimal  places  in  both  factors  ;  and  if  there  be  not  so  many  in 
the  product,  supply  the  deficiency  by  prefixing  ciphers. 


Multiply  ,02345  by  .00163. 

.02345 
.00163 


7035 
14070 
23  i  5 


.0000382235  Product. 


106  ARITHMETIC. 


2.  Multiply  54.25  by  2.2S02.  Ans.  123.70536. 

3.  Multiply  15.278  by  .7854.  Ans.  12,  iienilv. 

4.  Multiply  79.317  by  23.15,  Ajis.  1836.88305. 

5.  Multiply  350  by  .7853.  Ans.  274.855. 
fi.  Multiply  20  by  .55.  Ans.  \l. 

7.  Multiply  one-tenth  by  one-tenth.  A7is.  ,01. 

8.  Multiply  25  by  twenty-live  hundredths,  Ajis.  6.25. 

9.  At  4^  mills  apiece,  what  will  400  quills  cost  ?       Aiis.  ^1.80. 

To  multiply  decimals  by  10,  100, 1000,  Sfc,  remove  the  decimal 
paint  as  many  places  to  ike  right  as  there  are  ciphers  in  the  multi- 
plier, agreeably  to  Article  13.  Removing  the  decimal  point  one  fig- 
ure to  the  right  increases  every  figure  ten-fold,  &c. 

10.  Multiply  4.56  by  10.  Ans.  45.6. 

11.  Multiply  §1.75  by  100.  Ans.  $175. 

12.  Multiply  .006  by  1000.  Afis.  6. 

13.  A  benevolent  person,  whose  income  was  $6000,  gave  .12  of  it 
for  charitable  purposes  ;  what  did  he  give  away  ?  Ans.  $720. 

14.  The  capital  stock  of  a  bank  was  |  of  a  million  of  dollars,  and 
_3_  of  it  was  owned  equally  by  4  individuals  ;  how  much  was  owned 
by  each  ?  Ans.  $22500. 


DIVISION   OF  DECIMALS. 

(AiiT.  61.)  Division  being  the  reverse  of  multiplication,  and 
it  having  been  shown,  in  multiplication  of  decimals,  that  the  product 
of  any  two  factors  must  contain  as  many  decimal  places  as  the  num- 
ber of  decimals  in  both  factors  :  now,  the  divisor  and  quotient,  in  divi- 
sion, may  be  considered  as  factors  of  the  dividend ;  hence,  the  number  of 
decimal  places  in  the  quotient  must  be  equal  to  the  difference  of  the 
number  of  places  in  the  divisor,  and  the  number  made  use  of  in  the 
dividend. 

If  the  dividend  really  have  no  decimals,  make  a  decimal  point  at 
the  right  of  the  unit,  and  fill  up  the  blank  decimal  places  with  ciphers, 
as  many  as  you  wish.  In  dividing,  the  decimal  places  may  run  out  ; 
we  may  then  annex  ciphers  and  count  them  as  decimals.  If  the  dividend 
has  decimals,  but  not  so  many  as  the  divisor,  fill  up  blank  places,  as 
before,  which  will  not  affect  the  value  of  the  decimal,  (Art.  57). 
From  the  above  explanations,  the  reason  of  the  following  rule 
must  be  evident: 

Rule.  Divide  as  in  whole  numbers,  and  poiyit  off  from  the  right 
of  the  quotient  as  many  places  for  decimals  as  the  decimal  places  in 
the  dividend  exceed  those  in  the  divisor. 


DECIMAL    FRACTIONS. 

107  1 

1. 

H 

2de( 
more 
made 
two 

2 

3. 

4. 

5. 

fi. 

7. 

EXAMPLES. 

Divide  42.42  by  2.113. 

Operation. 
2.113)42.4200(20.07 
42.26 

ICOOO 
14791 

^ere  were,  originally, 

oration  we  used  three 

s  with  ciphers.    This 

taken  from  it,  leaves 

Quo^.O.llI. 

Quot.  11  1. 

Quot.  5.4232. 

Quot.  54.232. 

Quot.  16.278. 

30  by  .03. 

ere  we  rest,  and  count  off  the  quotient.     T 
:imal  places  in  the  dividend,  and  in  the  op( 
;  that  is,  filled  up  three  blank  decimal  place 
}  five,  and  three,  the  number  in  the  divisor 
decimal  places  in  the  quotient. 

Divide  2.3421  by  21.1. 

Divide  2.3421  by  .211. 

Divide  .8297592  by  .153. 

Divide  8.297592  by  .153. 

Divide  12  by  .7854. 

Divide  3  by  3  ;  divide  3  by  .3 ;  3  by  .03  ; 

(Art.  62.)  The  preceding  rule  for  division  of  decimals  is  as  clear 
and  as  explicit  as  any  mere  ruJe  can  be  :  but,  so  careless  are  pupils 
in  general,  about  the  decimal  point,  that  little  or  no  reliance  can  be 
placed  on  the  value  of  their  quotients,  until  they  are  taught  to  use 
iheir  judgments  in  each  and  every  case,  independent  of  any  rule. 
To  call  into  exercise  this  individual  reliance,  is  the  object  of  the  fol- 
lowing remarks  and  explanations. 

To  divide  —  to  cut  into  parts  —  will  not,  at  all  times,  give  a  clear 
understanding  of  division,  and  confusion  frequently  arises  from  taking 
this  view  of  the  subject;  we  better  consider  it  as  one  number  mea- 
suring another.  For  example  :  How  often  will  .5  of  a  foot  measure 
12  feet?  In  other  words,  divide  12  by  .5,  or  divide  12  by  ^.  Here, 
if  the  student  should  imagine  that  12  must  be  cut  into  parts,  he  would 
make  a  great  error.  He  must  divide  120  tenths  into  parts  ;  in  this 
cai^e,  into  5  parts,  because  the  5  is  .5 :  or,  he  may  consider  that  ^  of 
a  font  may  be  laid  down  in  12  feet ;  that  is,  measure  12  feet  24  times. 
Or,  he  may  reduce  the  12  feet  to  half  feet,  and  then  divide  by  1.  In 
fill  cases,  the  divisor  and  dividend  must  be  of  the  same  denomination, 
before  the  division  can  be  effected.  But,  in  decimals,  those  rcductitms 
are  made  so  easily  that  a  thoughtless  operator  rarely  perceives  them ; 
hence  the  difficulty  in  ascertaining  the  value  of  the  quotient. 

We  now  give  a  few  examples,  for  the  purpose  of  teaching 
the  pupil  how  to  use  his  judgment.  He  will  then  have  learned 
a   rule  more   valuable   than   all  others. 


108  ARITHMETIC. 


EXAMPLES. 

DiviJe  15..34  by  2.7.  Here,  we  consider  the  whole  number,  1.5,  is 
to  be  divided  by  less  than  3 :  the  quotient  must,  therefore,  be  a  Httle 
over  5.  One  figure,  then,  in  the  quotient,  will  be  whole  numbers,  the 
rest  decimals. 

Divide  15.34  by  .27.  Here  we  perceive  that  15  is  to  be  divided,  or 
rather  measured,  by  less  than  |  of  1  ;  therefore,  the  quotient  must  be 
more  than  3  times  15.  Or,  we  may  multiply  both  dividend  and  di- 
visor by  100,  which  will  not  aflect  the  quotient,  and  then  we  shall 
have  1534  to  be  divided  by  27.  Now,  no  one  can  mistake  how 
much  of  the  quotient  will  be  whole  numbers :  the  rest,  of  course, 
decimals. 

Divide  45.30  by  .015.  Conceive  both  numbers  to  be  multiplied  by 
1000  ;  then  the  requirement  will  be  to  divide  45300  by  15,  a  common 
example  in  whole  numbers. 

By  attention  to  this  operation,  the  student  will  have  no  difficulty  in 
any  case  where  the  divisor  is  less  than  the  dividend. 

Here  is  one  of  the  most  difficult  cases  : 

Divide  .003753  by  625.5.  In  all  such  examples  as  this,  we  insist 
upon  the  formality  of  placing  a  cipher  in  the  dividend,  to  represent 
tlie  place  of  whole  numbers,  thus : 

625.5)0.0037530( 

V/e  now  consider  whether  the  whole  number  in  the  divisor  will  be 
contained  in  the  whole  number  in  the  dividend,  and  we  find  it  will 
not :  we,  therefore,  write  a  cipher  in  the  quotient,  to  represent  the 
place  of  whole  numbers,  and  make  the  decimal  on  the  right,  thus :  0. 

We  now  consider  that  625  will  not  go  in  the  ID'S,  nor  in  the  lOO's^ 
nor  in  the  3,  nor  in  the  37,  nor  in  the  375,  but  it  will  go  in  the  3753^ 

We  must  make  a  trial  at  every  step,  that  is,  every  time  we  take  in 
vieiv  another  place ;  and  we  must  take  but  one  at  a  time.  In  this 
case,  then,  we  shall  have  0.000006,  the  quotient. 

Divide  3  by  30.  30  will  not  go  in  3  ;  we,  therefore,  write  0  for 
place  of  whole  numbers,  and  then  say,  30  in  30  tenths,  1  tenth 
times,  or  0.1. 

Divide  .55  by  11. 

11)0.55(0.05. 

1 1  in  0,  no  times ;  1 1  in  5  tenths,  no  times  ;  1 1  in  55  hundredths, 
5  hundredths  times.  It  will  be  observed  that  we  make  the  decimal 
point    in    the    quotient  as  soon  as   we  ascertain  it;  not  wait,  and 


DIVISION    OF    DKCIMALS.  109 


then  find  where  it  should  be,  by  counting,  &c., — ;i  rule 
against  which  we  have  notliing  to  say;  but  good  jutig- 
ment,  at  ready  command,  is  far  superior  to  any  rule  that 
can  be  formed. 

EXAMPLES* 

1.  Divide  9  by  450.  ^ns.  0,02. 

2.  Divide  2,39015  by  ,007.  J9ns.  341,45. 

3.  Divide  100  by  ,25.  ^ns.  400,00. 

4.  If  350  pounds  of  beef  cost  $12,25,  what  is  the  cost 
of  one  pound  ?  *^ns.  ,035. 

5.  At  $5,75  per  yard,  how  much  cloth  can  be  pur- 
chased with  $19,40625?  ^ns.  3,375  yards. 
^-  6.  At  7  per  cent.,  how  much  capital  must  be  invested 
to  yield  602  dollars  ?  ^ns.  $8600. 

7.  If  a  contribution,  amounting  to  $36,72,  be  made  by 
a  congregation  consisting  of  918  persons,  how  much  is  it 
a-piece?  ^ns.  ,04,  or  4  cents. 

8.  If  275  lemons  cost  $2,475,  how  much  is  it  a-piece? 

^ns.  9  mills. 

9.  A  benevolent  individual  gave  away  $600  per  annum 
to  charitable  objects,  which  was  ,12  of  his  income  ;  what 
was  his  income  ? 

(Art.  63.)  By  paying  some  litde  attention  to  the  rela- 
tion of  numlDcrs,  we  may  abbreviate  certain  divisions,  in 
a  similar  manner  as  we  abbreviated  multiplication,  m 
Art.  21. 

To  divide  by  ,25  is  the  same  as  to  multiply  by  4. 
"       "      by  ,5  is  the  same  as  to  multiply  by  2. 
"       "      by  ,75  is  the  same  as  to  multiply  by  4,  and 
divide  by  3. 

To  divide  by  ,125  is  the  same  as  to  multiply  by  8. 
"       "      by  ,333|-  is  the  same  as  to  multiply  by  3. 
"       '«       by  ,666|   is  the  same  as  to  multiply  by  3, 
and  divide  by  2,  Slc.  &c.,  for  any  aliquot  part  of  unity. 
The  same  is  true  of  aliquot  parts  of  10,  100,  1000,  &c. 

EXAMPLES. 

1.  Divide  6  by  ,25.  ^ns.  24. 

2.  Divide  6  by  25.  -       Jns.  ,24. 

3.  Divide  15  by  16|,=^  of  100.  ^ns.  ,9. 


110  ARITHMETIC. 

4.  Divide  15  by  ,16|.  ^ns.  90. 

5.  Divide  4,27  by  ,33^.  .^ns.  12,81. 

REDUCTIOIN   OF  DECIMALS. -.-_ 

(Art.  64.)  To  reduce  a  vulgar  fraction  to  its  equiva- 
lent decimal :  the  value  of  any  fraction  is  found  by  di- 
viding the  numerator  by  the  denominator;  thus,  |=2. 
But  in  a  proper  fraction,  we  cannot  divide  the  numerator 
without  first  reducing  it.  Annexing  one  cipher,  makes  it 
tenths ;  two  ciphers,  hundredths ;  and  so  on.  The  quo- 
tient, therefore,  will  be  decimal  parts.  Thus :  reduce  | 
to  a  decimal. 

8)1,000 

,125 

We  may  take  this  view  of  the  case :  Multiply  both 
numerator  and  denominator,  by  10,  100,  1000,  &c. ; 
which  will  not  change  the  value  of  the  fraction,  (Art.  35.) 
We  then  have  |^^^.  Now  divide  both  numerator  and 
denominator  by  8,  and  we  have  yVVo'  which  is  a  deci- 
mal fraction,  by  the  definition  of  decimals.  Hence, 
whichever  view  we  take,  we  have  the  following 

Rule.  Place  a  decimal  point  at  the  right  of  the  nu- 
merator, annex  ciphers,  and  divide  by  the  denominator. 
Compound  fractions  must  first  be  reduced  to  simple 
fractions. 

EXAMPLES. 

1.  Reduce  \,  \,  and  |,  to  equivalent  decimals. 

Jins.  ,25,  ,5,  ,75. 

2.  What  decimal  is  equivalent  to  |?         Ans.  ,625. 

3.  What  decimal  is  equivalent  to  |  of  |  ?    Ans.  ,4. 

4.  Reduce  |f  to  a  decimal.  Ans.  ,9375. 

5.  Reduce  y^'jij  to  a  decimal.  Ans.  ,0008. 

6.  Reduce  jf  to  a  decimal.  Ans.  ,7058  + 

There  are  many  vulgar  fractions  that  cannot  be  exactly 
expressed  in  decimals.    The  following  are  some  of  them : 


2       1       2       4.     A      _1_      _2_      Stjf, 


REDUCTION    OF    DECIMALS.  HI 


Reduce  I  to  a  decimal.  Ans.  ,3333,  <fec.  This  is 
called  a  repeating,  or  infinite  decimal. 

Reduce  /y  to  a  decimal.  Ans.  181818,  &c.  This  is 
called  a  circulating  decimal,  and  the  repeated  figures  are 
called  a  compoinid  rej)etend.  Compound  repetends  may- 
consist  of  three  or  more  figures — as,  ,123  123,  &c. ;  or, 
,134  91349,  &c. 

Reduce  ^"3  to  a  decimal.  Ans.  ,121212  -|- 

(Art.  65.)  To  reduce  a  decimal  to  a  vulgar  fraction. 

Rule.    Write  the  dejiominator  under  the  decimal, 
and  reduce  to  its  lowest  terms. 
This  rule  requires  no  comment. 

EXAMPLES. 

1.  Reduce  ,75  to  a  vulgar  fraction.      Ans.  //^  =  |. 

2.  Reduce  ,125  to  a  vulgar  fraction.  Ans.  \. 

3.  Reduce  ,9375  to  a  vulgar  fraction.  Ans.  i-f. 

4.  Reduce  ,16  to  a  vulgar  fraction.  Ans.  ~j. 

5.  Reduce  ,15  to  a  vulgar  fraction.  Ans.  t,\. 

6.  Reduce  ,124  to  a  vulgar  fraction.  Ans.  ^\. 

7.  Reduce  ,655  to  a  vulgar  fraction.  Aiis.  ^fi- 

(Art.  66.)  To  reduce  a  repetend,  or  circulating  deci- 
mal, into  its  equivalent  vulgar  fraction.  By  a  little  in- 
spection of  the  fractions  expressed  in  Art.  64,  we  per- 
ceive that  the  denominators  of  fractions  which  make  re- 
petends, are  9,  11,  or  factors  of  9,  11.     Hence  this 

Rule.  Make  the  repetend  the  num,erator,  and  for  the 
denominator  take  as  many  9's  as  there  are  figures  in 
the  repetend. 

EXAMPLES. 

1.  Reduce  ,6  to  a  vulgar  fraction.  Ans.  |. 

2.  Reduce  12.3  to  a  vulgar  fraction.  Ans.  ~^^. 

3.  Reduce  ,18,  &c.  to  a  vulgar  fraction. 


Ans.  U--=j\. 


We  shall  resume  this  subject  again,  when  we  treat  of 
geometrical  progression. 


112  ARITHMETIC. 


(Art.  67.)  To  find  the  value  of  a  decimal  in  known 
parts,  of  integers  of  inferior  denominations. 

Rule.  To  reduce  a  decimal  from  a  higher  to  a  lower 
denominaiion :  we  reason  and  mnlliply  just  as  we 
would  if  the  sum  to  be  reduced  was  a  whole  number,  in 
place  of  a  decimal^  only  observing  the  decimal  point. 

EXAMPLES. 

What  is  the  value,  in  shillings,  pence,  &:c.,  of  ^66?  or 
any  other  number  of  pounds  ?  We  should  certainly  mul- 
tiply by  20,  12.  &c.;  and  whatever  we  do  to  one  or  more 
pounds,  we  should  do  to  apart  of  a  pound. 

1.  What  is  the  value  of  ,6237  of  a  pound  sterling,  ex- 
pressed in  lower  denominations  ? 
,6237 
20 


12,4740= the  shillings  and  decimals  of  a  shilling,  in  the 
12         decimal  ,6237  of  a  pound. 


5,6880 =the  pence  and  decimals  of  a  penny,  in  the  de- 
4         cimal  ,4740  of  a  shilling. 


2,7520 =the  farthings  and  decimals  of  a  farthing  in  the 
decimal  ,6880  of  a  penny. 
The  integers  on  the  left  of  the  decimal  points,  are  the 
numbers  that  compose  the  answer. 

^ns.  12  s.  5  d.  2}  farthings. 

2.  What  is  the  value  of  ,725  of  a  £  ? 

^ns.  15  s.  6d. 

3.  What  is  the  value  of  ,8285  of  an  acre  ? 

^^ns.  3  roods,  12,56  poles. 

4.  What  is  the  value  of  ,432  of  a  hundred  weight  ? 

.^ns.  Iqr.  20  lb.  6oz. + 

5.  What  is  the  value  of  ,234  of  a  day  ? 

^7is.  5  h.  36  m.  57y\  sec. 

6.  How  many  furlongs,  &:c.,  in  ,784  of  a  mile  ? 

.^ns.  6  fur.  10  rods,  4  yds.  2  ft.  -|- 

7.  How  many  quarters,  &.C.,  in  ,936  of  an  ell  Eng- 
lish ?  ^7is.  4  qr.  2  n.  + 


REDUCTION    OF    DECIMALS.  113 


(Art.  08.)  The  reverse  of  Art.  67.  '  To  reduce  low- 
er denominations  to  decimals  of  a  higher.  As  higher 
is  reduced  to  lower  by  midtip/icafiou,  lower  must  be  re- 
duced to  higher  by  division,  beginning  with  the  lowest 
denomination.     These  observations  explain  the  following 

Rule.  Divide  each  denomination,  beginning  with 
the  lowei^t,  by  such  a  number  as  will  bring  it  to  the  next 
superior  denomination,  and  set  the  quotient  on  the  right 
of  that  denomination,  with  a  decimal  point  betiveen 
them.  Then  divide  the  next  higher,  together  with  its 
decimal,  by  the  number  that  will  bring  it  to  its  next 
higher  denomination,  in  the  same  manner  as  before^ 
and  so  on  as  far  as  necessary. 

EXAMPLES. 

1.   What  part  of  a  hundred  weight  is  3  qr.  14  lb.  ? 

Jlns.  ,875. 
3Iore  circuitously.  281 14, 


First,  (by  Art.  46)  reduce  to  a  vulgar  fraction,    4 1   3,5 


0,875 
Then  (by  Art.  64)  the  vulgar  to  a  decimal,  thus : 

3  qr.  14  lb. =98  pounds.  Fraction    -' «  -  875 

1  cwt.  =  112  pounds.  i-iaction,  tt2-»«75. 

2.  Reduce  2  quarters  3  nails  to  the  decimal  of  a  yard. 

Ans.  ,6875  yd. 

3.  What  part  of  a  bushel  is  3  pecks  1,12  quarts  ? 

Ans.  ,785  bu. 

4.  What  part  of  1  hogshead  is  1,2  pints? 

Ans.  ,00238  hhd. 

5.  Reduce  5  furlongs,  12  poles,  to  the  decimal  of  a  mile. 

Ans.  ,6625. 

6.  Reduce  55  minutes,  37  seconds,  to  the  decimal  of 
a  day.  Ans.  ,03862  -|- 

7.  Reduce  ,32  of  a  pint  to  the  decimal  of  a  bushel. 

Ans.  0,005. 

8.  Reduce  10  ounces,  13  pennyweights,  9  grains,  to  the 
decimal  of  a  pound  Troy.  Ans.  ,8890625. 

If  more  examples  are  desired,  reverse  those  under 
Art.  67. 

k2 


114  ARITHMETIC. 


Astronomical  Problems. 
The  mean  length  of  a  tropical  year  is  365  days,  5  hours, 
48  minutes,  49  seconds ;  what  decimal  of  a  day  over  365  ? 

Ans.  ,24223  + 
The  tropical  revolution  of  the  moon  is  27  days,  7  hours, 
43  minutes,  3  seconds;  how  many  days,  and  decimals 
of  a  day?  e^ns.  27,321566+ 

MISCELLANEOUS    EXAMPLES, 

Applicable  to  the  various  rules  infractions,  both  vulgar 
and  decimal. 

1.  Which  is  greatest,  V|,  or  f  |  ? 

2.  Which  is  most,  |  of  a  rod,  or  |  of  12  feet? 

3.  If  the  dividend  be  ^,  the  quotient  320,  what  is  the 
divisor  ? 

4.  The  divisor  being  \,  the  quotient  y\  ;  what  is  the 
dividend  ? 

5.  The  divisor  being  4520,  the  dividend  | ;  what  is 
the  quotient? 

6.  The  Ci\  isor  to  a  number  is  ,02,  the  quotient  is  420; 
what  is  the  number  ? 

7.  What  is  the  difference  between  f  of  £T,  and  i-  of 
a  shilling? 

8.  What  is  the  sum  of  19  and  5  hundredths  ;  120  and 
9  thousandths ;  36  ten-thousandths  ;  and  406  and  eight- 
millionths  ? 

9.  From  yf^^  take  y^'/o  u'  and  divide  the  difference 
by  ,002.  ^'ins.  37,7. 

10.  The  sum  of  $120  36  was  to  be  paid  by  a  certain 
number  of  persons,  and  it  amounted  to  3  dollars  9  mills 
a-piece ;  how  many  were  there  of  them  ?       Ans.  40. 

11.  Add  I  of  a  yard  to  -^  of  a  foot.  Ans.  1  yd. 

12.  Add  f  of  a  day  to  |  of  an  hour. 

Ans.   16  h.  40  m. 

13.  Add  I  of  a  week,  ^  of  a  day,  and  \  of  an  hour. 

Ans.  1  d.  16  h.  36  m. 

14.  At  31  i  cents  per  bushel,  how  many  bushels  can  be 
purchased  for  9  dollars  ?  Ans.  28, 8^ 

15.  At  62^^  cents  per  basket,  how  many  baskets  can  be 
bought  for  16  dollars  ?  Ans.  25,6. 


REDUCTION    OF    DECIMALS. 


115 


141 


16.  How  many  yards  of  cloth  in  4  remnants  contain 
ing,  severally,  3  yards  3  quarters,  2  yards  3  nails,  Ij 
yards,  and  2^  yards  I  .^ins.  10,4375. 

47,  Trom  1  hundred  weight,  subtract  3  quarters 
pounds.  ..^?2s.  ,11830  cwt. 

18.  What  is  the  cost  of  3f  yards  of  cloth,  at  5|  dollars 
per  yard?  c^?i5.  $19,406 + 

19.  At  5|  dollars  per  yard,  how  much  cloth  can  be  pur- 
chased with  19,40625  dollars?  .tljis.  3,375  yd. 

2Q-.  At  11,76  dollars  per  hundred  weight,  what  will  i 
quarter  of  sugar  come  to  ?  ^ns.  $1,47. 

21.  Traveling  at  the  rate  of  4^-  miles  an  hour,  in  how 
many  hours  will  a  foot-man  go  34|  miles  ? 

*-%is.  7,5  hours. 

22.  If  85  yards  of  cloth  be  bought  for  191,25  dollars, 
and  sold  at  2  dollars  87 ^^  cents  per  yard,  how  much  is 
the  whole  profit  ?  "  c^??.s.  $53,12'^. 

23.  A  person  bought  3^  pounds  of  tea,  at  1,125  dollars 
per  pound;  8  pounds  of  coffee,  at  12^  cents;  4  gallons 
of  molasses,  at  31i  cents  per  gallon;  20  pounds  of  su- 
gar, at  8V  cents  per  pound;  what  is  the  amount  of  the 
bill?        "  ./^nv.  $7,886 -f- 

24.  If  4i-  gallons  of  wine  cost  5-'  dollars,  what  was  it 
per  gallon?  ~j9ns.  $1,27,  nearly. 

25/'  Two  men  start  on  a  journey  at  the  same  time;  one 
travels  at  the  rate  of  41^  miles  per  hour,  the  other,  5| ; 
how  many  hours  after  their  departure  will  they  be  34 
miles  apart,  and  how  far  will  the  most  advanced  have 
traveled?  Jins.  In  24  hrs. ;  dis.  136  miles. 


116  ARITHMETIC. 


SECTION  IV 


COMPARISON  OF  NUMBERS. 

(Art.  69.)  A  number  may  be  considered  as  part  of 
another  number;  for  instance,  3  is  one-half  of  6,  and  2 
is  one-fourth  of  8  ;  and  we  may  mechanicaUy  find  the 
V  and  i,  by  dividhig  the  first  number  by  the  second ; 
thus,  f  =  4-:  f  =  !r'  Or,  the  comparison  remains  the 
same,  if  we  divide  both  numbers,  (the  numbers  we  com- 
pare,) by  the  smallest  number;  thus,  3  compares  with  6, 
as  1  to  2 ;  2  to  8,  as  1  to  4. 

What  part  of  3  is  2?  Certainly  f.  What  part  of  2 
is  3  ?     Certainly  |. 

Now,  wherever  this  phraseology  applies,  (JJliat  pari 

of is,)  the  number  after  the  word  is,  must  be 

taken  for  a  numerator,  and  the  other  for  a  denominator; 
and  reduce  the  fraction  to  its  most  simple  form,  and  we 
shall  have  the  most  simple  relation  of  one  number  to  the 
other.     (See  Art.  46.) 

EXAMPLES. 

1.  What  part  of  7  is  2  ?  Ans.  f .  That  is,  7  is  to  2, 
as  1  to  i. 


2.  What  part  of  12  is  4  ? 
to  4,  as  1  to  |. 

3.  What  part  of  19  is  7? 

JJns.  |.     ( 

3r,  we  may  say 

4.  What  part  of  7  is  19  ? 

./Ins.  y. 

5.  What  part  of  f  is  ^  ? 

Ans.  1. 

6.  AVhat  part  of  f  is  f  ? 

Ans.  |i. 

Make  complex  fractions  of  the  two  last  examples,  and 
reduce  them. 

(Art.  70.)  Before  numbers  can  be  compared  in  this 
way,  they  must  be  of  the  same  denomination.  If,  for 
example,  we  would  compare  5  feet  with  2  yards,  and 
asked  what  part  of  five  is  two  we  could  not  say  ^  ;  for 
we  nil  know  that  2  yards  is  more  than  five  feet.     We 


COMPARISON    OF    NUMBERS.  117 


niiist  first  reduce  the  yards  to  feet;  and  then  the  question 
is,  What  part  of  5  feet  is  6  feet?  Jlns.  |. 

7.  What  part  of  3  hundred  weight  is  1  hundred  weight 
3  quarters  ?     Reduce  both  to  quarters  ;  then  compare. 

.fins.  -j-\. 

8.  What  part  of  6  shillings  is  8  shillings  6  pence  ? 

Jins.  f  1. 

9.  WHiat  part  of  4,6  is  ,46?  Ans.  j\. 

10.  Wliat  part  of  3  gallons  is  2  quarts  1  pint? 

1 1 .  Wh;it  part  of  1  yard  is  2  feet  6  inches  ? 

^718,  f . 

12.  What  part  of  5  dollars  is  35  cents  ?     Ans,  y|-^. 
Wliatpartof  12  is  101? 

What  part  of  3  is  f,-?" 

AVhat  partof  6\  is  6^  ? 

The  relation  of  numbers  and  things  compared  in  this 
■vvay,  may  be  applied  to  the  solution  of  a  great  variety  of 
practical  problems. 

Example.  If  3  men  can  build  7  rods  of  wall  in  a  day, 
how  many  rods  can  9  men  build  ?  Here  we  compare  9 
men  with  3  men  ;  and  we  say,  what  part  of  3  is  9  ?  A7is. 
3  times.  Now,  the  rods  of  wall  must  compare,  as  the 
men  that  build  them.  Hence  7X3=21  rods  of  wall, 
for  the  answer  to  the  question. 

Example  2.  H  three  horses  eat  8  bushels  of  oats  in  2 
weeks,  how  long  will  it  take  them  to  eat  40  bushels  ?  8 
compares  to  40  bushels,  as  1  to  5;  hence  2X5=10 
weeks. 

Example  3.  If  4'^  tons  of  hay  keep  3  cattle  over  the 
w^inter,  how  many  tons  will  it  require  to  winter  25  cattle? 
3  cattle  compare  with  25,  as  1  to  Y;  hence,  V  x|  =  y 
=375  tons,  Jlns. 

Example  4.  If  19  gallons  of  molasses  cost  12  dollars, 
what  w^ill  3^  quarts  cost?  Here  we  cannot  directly 
compare  gallons  with  quarts;  one  or  the  other  must  be 
reduced.  The  quarts  are  ^-^ .  Rut  to  reduce  quarts  to 
gallons,  we  divide  by  4;  therefore,  our  3 J  quarts  is  |^, 
expressed  in  gallons.  Now,  19  compares  with  \'l,  as  1 
to  ^^5  hence  12XaV=A  dollars,  or  60  cents  the  Aiis. 


118  ARITHMETIC. 


Example  5.  If  16  busliels  of  oats  cost  $6,75,  what 
will  320  bushels  cost?  Oats  compares  with  oats,  in  the 
supposition  and  demand  ;  as,  16  to  320  ;  or,  as  1  to  20. 
Hence,  6,75X20  =  135  dollars.  Am. 

Example  6.  If  1  acre  and  20  rods  of  ground  produce 
45  bushels  of  wheat;  at  that  rate,  how  much  will  9  acres 
produce  ?  One  acre  and  20  rods  is  180  rods.  Nine 
acres  is  9X  160  rods.  180  compares  with  9X  160.  As 
20  compares  with  160,  by  dividing  both  by  9  ;  or,  as  1 
to  8,  by  dividing  both  by  20.  Hence,  45  X  8=360  bush- 
els, the  Ans. 

Example  7.  If  f  of  a  yard  of  cloth  be  worth  4  dollars, 
what  will  be  the  worth  of  ^  of  a  yard  ?  Observe,  f =|. 
Compare  |  to  ^,  which  is  the  same  as  6  to  7.  But  6  is 
to  7  as  1  to  \.     Hence  ^X4=4f  dollars,  Ans. 

Example  8.  If  3  yards  of  cloth  cost  12|  dollars,  how 
many  yards  may  be' bought  for  102  dollars?  Observe, 
12|  =  \';  Y  is  to  '^2  as  1.  to  |.  But  i  is  to  2  as  1  to 
8.  '  Hence  3  X  8=24  yards,  the  Ans. 

9.  Boarding  at  12  shillings  6  pence  per  week,  how 
long  will  32  pounds  10  shillings  last  me  ? 

Ans.  52  weeks. 

10.  If  a  barrel  of  beef  last  10  men  95  days,  how  long 
will  it  last  25  men  ?  Ans.  38  days. 

11.  If  I  lend  a  man  $200  for  60  days,  how  long  ought 
he  to  lend  me  $275  to  requite  the  favor  ? 

Ans.  43yy  days. 

12.  How  many  years  win  it  require  for  5  cents  to  gain 
the  same  interest  that  $100  does  in  1  year  ? 

Ans.  2000  years. 

13.  If  it  be  required  to  line  cloth  i  of  a  yard  wide, 
with  lining  |  wide,  what  must  be  the  relative  quantity  of 
the  lining?  Ans.  ^r. 

14.  If  22  cents  buy  3  pounds  of  coffee,  how  many 
pounds  will  44  dollars  buy  ?  Ans.  600. 

15.  If  3  pounds  are  purchased  for  25  cents,  what  will 
135  pounds  cost?  Ans.  $11,25. 

16.  At  41  dollars  per  ton,  how  many  ton  can  be  pur- 
chased for  $900  ?  Ans.  200. 


PROPORTION.  119 


PROPORTION. 

(Art.  71.)  From  the  comparison  of  numbers  we  de- 
rive proportion,  one  of  the  most  fruitful  and  valuable 
principles  in  arithmetic,  tlie  mathematics,  and  natural 
philosophy.  Too  much  cannot  be  said  of  its  importance 
and  utility. 

Definitions. 

1.  When  two  numbers  are  directly  compared,  by  di- 
viding one  by  the  other,  as  in  Article  69,  the  quotient 
expresses  their  relation. 

2.  If  two  other  num'ners  are  compared  in  the  same 
way,  and  we  find  the  same  quotient  the  saine  relation, 
the  four  numbers  may  constitute  a  proportion. 

3.  Two  numbers,  thus  compared,  are  called  a  couplet. 

4.  Tiie  quotient,  or  relation,  is  technically  called  the 
ratio. 

For  example  :  If  we  compare  2  to  4,  by  dividing  4  by 
2  we  find  the  intio,  2.  Also,  3  compares  with  6  by  the 
same  ra'io.     Therefore,  2  is  to  4  as  3  is  to  6. 

'i'he  words  between  the  terms  need  not  be  written,  as 
points,  thus,  :  : :  :  or,  :  =  :  express  the  same  as  the 
written  words,  (Definition,  Art.  5).     Hence, 

2     :     4     :  :     3     :     6, 

is  a  proportion.  The  ratio  to  this  proportion  is  2.  Ob- 
serve, that  the  product  of  the  extreme  terms  is  equal  to 
the  product  of  the  mean  or  middle  terms  ;  that  is  2X6 
=4X3,  and  this  is  true  in  every  proportion.  If,  in  any 
proposed  case,  this  is  not  true,  the  proportion  is  false, 
and  is  not,  in  fact,  a  proportion. 

(Art.  72.)  This  property  of  extremes  and  means, 
results  from  the  equality  of  the  ratio  between  the 
couplets. 

Thus, 2:4,  i  =  ratio  2. 

3  :  6,  |=ratio  2. 
Hence,  {=|,  because,  things  equal  to  the  same  thing 
equal  one  another. 


120  ARITHMETIC. 


6X2 
Multiply  both  terms  by  2,  and  4= .     Now  mul- 

o. 

tiply  by  3,  and  3X4=6X2  ;  that  is,  the  product  of  the 
extremes  equals  that  of  the  means.  12  :  48  compare,  as 
1  to  4;  that  is,  the  ratio  between  the  terms  of  this  cou- 
plet is  4.*  3  :  12  compares,  as  1  to  4,  same  ratio  ;  there- 
fore 12  :  48  ::  3  :  12  must  be  a  perfect  proportion ;  hence, 
12X12=48X3. 

(Art.  73.)  The  terms  of  this  proportion  may  be  chang- 
ed, and  still  constitute  :i  perfect  proportion  ;  but  will  have 
a  ditl'erent  ratio  between  the  couplets. 

1.  Invert  the  means,  and  12  :  3  ::  48  :  12.  The  ra- 
tio is  now  \ ;  but  the  product  of  the  extremes  and  means 
is  the  same  as  before. 

2.  If  we  treat  both  couplets  exactly  alike,  no  matter 
what  we  do,  the  result  will  still  be  a  proportion.  Let 
us  subtract;  the  second  from  the  first  is  to  the  2d,  as  the 
fourth  from  the  third  is  to  the  4th ;  that  is,  9  :  3  ::  36  :  12. 
These  couplets,  we  perceive,  have  the  equal  ratio,  ~,  and 
the  product  of  the  extremes  is  equal  to  the  product  of  the 
means.  Again :  the  second  from  the  first  is  to  the  first, 
as  the  fourth  from  the  third  is  to  the  second.  That  is, 
9  :  12  ::  36  :  48.  V=|-  ^atio ;  Also,  ff=J  ratio. 
Hence  this  proportion  is  true. 

(Art.  74.)  Two  or  more  proportions  may  be  multi- 
plied together,  term  by  term,  and  constitute  a  new  pro- 
portion. 

Thus,  as 2:4::       5:10 

Also, 6  :     1  ::    24  :    4 

and 9:3::       3:1 


Then, 108  :  12  ::  360  :  40 


*  Some  mathematicians,  particularly  the  English,  reduce  the  se- 
cond term  to  unity,  by  dividing  both  terms  by  the  second.  In  that 
case,  the  ratio  of  12  to  48,  would  be  i,  in  place  of  4.  Either  way 
is  correct,  if  we  are  uniform  and  consistent.  We  prefer  the  French 
method,  of  reducing  the  first  term  to  unity,  conceiving  it  more  sim- 
ple than  the  other. 


PROrORTION.  121 


Vt'e  may  divide  the  first  couplet  by  12,  the  second  by 
40,  and  we  have  9  :  1  ::  9  :  1,  a  proportion  manifestly 
true.  We  may  multiply  or  divide  either  antecedents  or 
consequents  by  the  same  number,  and  we  still  have  a 
proportion. 

(Art.  75.)  By  retaining  strong  hold  of  the  fact,  that 
the  products  of  the  extreynes  and  means  arc  equal,  in 
every  geometrical  proportion,  we  can  find  any  term,  or 
factor  of  a  term,  that  may  be  wanting  or  lost.  Take 
the  proportion,  2  :  4  ::  5  :  10.  If  the  fourth  term,  the 
10,  be  wanting,  we  still  have  the  product  of  the  means. 
We  have  also  one  factor  of  the  extremes,  which,  used 
as  a  divisor  to  divide  the  common  product,  will  give  the 
other  factor,  or  the  fourth  term :  thus, 

4X5 

=10. 

2 

When  the  fourth  term  of  any  proportion  is  wanting,  we 
then  have  three  terms,  and  the  application  of  Proporion 
then  becomes  the  Rule  of  Three,  or  Golden  Rule.  The 
first  technicality  arose  from  the  fact  of  there  being  three 
given  terms  to  tind  a  fourth,  the  last  from  the  great  utilily 
and  beauty  of  its  application. 

Resume  the  proportion,  2  :  4  ::  5  :  10,  nnd  suppose 
the  second  term  wanting;  we  then  have  2  :  []  ::  5  :  10 
only.  The  extremes  now  give  us  the  common  product, 
20 ;  which,  divided  by  5,  gives  4,  the  term  wanting.  On 
this  principle,  if  only  one  term  is  wanting,  no  matter 
which  one,  it  can  be  found.  When  the  fourth  term  is 
wanting,  it  is  the  rule  of  three  direct.  When  the  fust 
term  is  wanting,  it  is  the  rule  of  three  inverse. 

(Art.  76.)  In  a  direct  proportion,  we  can  obtam  the 
fourth  term,  by  multiplying  the  third  term  by  the  ratio. 

Example.  5  :  15  ::  11  :  ,  to  the  fourth  term.  AVhat 
is'that  term?  5  :  15  are  as  1  to  3.  Then  3  is  the  ratio. 
Hence,  11X3=33,  Mns.  In  an  inverse  proportion,  the 
first  term  is  found  by  dividing  the  second  term  by  the 
ratio. 

Example.  []  :  50  ::  24  :  4.  What  part  of  24  is  4? 
Ans.  i=the  ratio.     Hence,  divide  50  by  ^=300  .fins. 


122  ARITHMETIC. 

Or,  in  consideration  of  the  common  product,  we  have, 

50X24 

for  the  first  term, =300  as  before. 

4 

(Art.  77.)  In  any  proportion,  suppose  300  :  50  ::  24  : 
4,  we  may  conceive  any  term  as  composed  of  two  or 
more  factors,  and  one  of  them  wanting.  It  can  be  re- 
stored by  means  of  the  common  product^  divided  by  the 
known  factors,  in  the  deficient  extremes  or  means,  as  the 
case  may  be.  Let  us  suppose  the  second  term,  50,  to 
be  separated  into  factors,  5X  10,  and  the  5  a  term  sought. 
We  then  have  300  :  []X  10  ::  24  :  4;  the  extremes  be- 
ing perfect,  gives  tlie  common  product.  The  means  are 
imperfect ;  but  if  we  divide  the  common  product  by  the 
factor  we  have  in  the  means,  the  quotient  will  give  the 
lost  or  sought  factor.  These  multiplications  and  divi- 
sions should  be  performed  as  directed  in  Art.  24. 


^6 
6  ^^ 


^^^   on  6  30 

^00  30  ^ 


5  Ans. 

If  we  have  300  :  50  ::  2X  []  :  4  :  to  find  the  lost 
factor  to  the  third  term,  the  process  is  the  same  as  be- 
fore. If  we  have  3X[]  :  50  ::  24  :  4,  to  find  the  lost 
or  sought  factor,  we  observe  that  the  means  are  now  per- 
fect, and  they  give  the  common  product,  and  the  ex- 
treme factors  are  divisors.     Thus: 

%. -^-^o^- 

APPLICATION  OF  THE  FOREGOING  PRINCIPLES. 

(Art.  78.)  Questions  which  should  fall  under  propor- 
tion, or  the  rule  of  three,  have  two  terms  of  comparison 
of  the  same  name  ;  one  in  the  supposition,  the  other  in 
the  demand.  The  third  term  is  of  the  same  name  as 
the  ansiver,  or  the  term  required.  To  state  a  question, 
is  to  arrange  the  terms  in  order,  agreeably  to  the  laws  of 
proportion.  When  the  question  is  direct — that  is,  more 
requiring  more,  or  less  requiring  less — state  the  question 
by  the  following 


PROPORTION.  123 


Rule.  Of  the  two  terms  of  comparison,  place  that 
of  supposition  first,  and  that  of  the  demand  second, 
and  that  of  the  same  name  as  the  answer  required,  the 
third.  Then  find  the  fourth  term  or  answer,  as  di- 
rected in  .irt.  75. 

Example  1.  If  12  bushels  of  wheat  are  wprth  IG  dol- 
lars, wliat  is  the  value  of  48  bushels  ? 

Here  are  two  terms  of  the  same  name,  (bushels,) 
which  we  call  terms  of  comparison,  and  one  term  (dol- 
lars) of  the  same  name  as  the  answer  required.  Of  these 
terms  of  comparison,  12  is  that  of  supposition.  We  know 
it  by  the  word  if,  or  one  of  like  import,  that  may  stand 
before  it,  the  48  is  that  of  demand,  as  the  language  most 
clearly  implies.     Hence,  by  the  rule, 

12  :  48  :  :   16  :  to  the  fourth  term. 

In  former  times,  when  mechanical  labor  was  consider- 
(h1  no  task,  we  would  multiply  48  by  16,  and  divide  the 
i  product  bv  12  ;  we  now  do  it  by  canceling,  according  to 
I  Art.  24. 

I       Or,  we  may  say,  12  to  48  as  1  to  4  ;  therefore  16X4 
j  =64  answer,  which  is  canceling  ivithout  the  form,  and 
j  which  is  generally  most  convenient  in  the  rule  of  three. 
Example  2.  If  7  pounds  of  sugar  cost  75  cents,  how- 
many  pounds  can  be  bought  for  9  dollars  ? 

Two  of  the  given  terms  in  this  question  are  of  the  same' 
kind  (money);  but  not  of  the  same  denomination.  These 
are  the  terms  of  comparison,  after  one  of  them  is  reduced 
to  the  denomination  of  the  other  (Art.  70). 

It  is  obvious  that  75  cents  is  the  term  of  supposition  ; 
hut  the  if,  (jpfssicn  cf  condition  is  not  directly  before  it ; 
however,  a  litde  change  in  the  phraseology  will  bring 
it  so. 

If  75  cents  M^ill  purchase  7  pounds  of  sugar,  how  ma- 
ny pounds  can  be  bought  for  9  dollars  ? 

Slate metit,  75  :  900  : :  7  :  to  the  answer. 
Divide  the  couplet  or  terms  of  comparison,  by  25, 

and 3   :  36  :  :  7  :  to  the  answer ; 

or, 1   :   12  :  :  7  :  84.  the  anwser. 


124  ARITHMETIC. 


$      $       lb. 
Same  question  again,  ^   :  9  :  :  7  :  the  answer. 

Divide  by  3 l-  :  3  :  :  7  :  the  answer. 

Mult'p'y.  by  4.  ...  1    :    12  ::  7  :  84  the  answer. 

As  a  specimen  of  the  manner  of  doing  things  in  for- 
mer days,  Kwe  cut  the  following  problem  and  solution, 
from  a  well-known  book. 

Example.  If  48  yards  of  cloth  cost  $67,25,  what  will 
144  yards  cost  at  the  same  rate  ? 
Operation, 
yd.      yd.  $ 

48  :   144  :  :  67,25  :  Ans. 
144 


26900 
26900 
6725 


48)9684,00($201,75 
96 


84 
48 

S60 
336 

240 
240 

Now  let  it  be  observed  once  for  all,  that  these  opera- 
tions are  those  of  comparison  or  proportion,  and  the  ac- 
tual numbers  given  need  not  be  used,  provided  we  use 
their  proportion. 

Here  48  is  to  144,  obviously,  as  4  to  12  ;  or,  as  1  to 
3.     Hence  67,25  X  3=$201,75,  Ans. 

(Art.  79.)  When  the  terms  are  of  various  denomina- 
tions of  weights,  measures,  &c.,  as  a  general  rule,  reduce 
them  all  to   the  lowest  denomination    mentioned    after 


F'ROPor.Tiox.  125 


statement ;  but  by  making  use  of  fractions,  we  may  very 
often  abridge. 

Example^.    If  3  pounds   12  ounces  of  tea  cost  $3,50, 
what  will  11  pounds  4  ounces  cost? 
lb.  oz.    lb.  oz.         $ 
Statement,.  •    3   12  :    11   4  ::  3,50 
In  place  of  reducing  the  Aveights  to  ounces,  we  observe 
12  0^.  =  ;;  lbs. 

lb.       lb.      money. 
^   :   \\\   :  :  3,50  :  Ans. 
Multiply  the  terms  of  comparison  by  4,  and 
15  :  45  :  :  3,50 

or, 1   :     3  :  :  3,50 

He^ce,  3,50X3= S10,50  Ans. 

The  following  problem  is  commonly  wrought  by  a 
very  tedious  process,  and,  indeed,  must  be,  if  rules  rath- 
er than  principles  are  to  be  our  guide. 

If  2  hundredweight  3  quarters  21  pounds  of  sugar 
cost  £6  1  s.  8  d.,  what  cost  35  hundred  weight  1  quar- 
ter ?  Ans.  £73. 

By  the  general  rules,  (which,  as  general  rules,  are  good,) 
we  must  reduce  the  quantities  to  pounds,  and  tlie  money 
to  pence,  either  after  or  before  the  statement. 

cwt.  qr.  lb.     cwt.  qr.  £.  s.  d. 

We  proceed  thus :    2     3     21   :  35     1     ::  .  .  .  6    1    8 

Reduce  to  qrs.  ..  11}    : 141  ::...6    1    8 

Multiply  by  4,  ..  47    141X4    ::...6    1    8 

Divide  by  47,  ...  1  .... ' 3X4 6    1    8 

Hence, 12 


Ans.  £73    0   0 
For  further  illustration,  we  extract  the  following  : 
If  7  pounds  of  coffee  cost  87^  cents,  what  must  I  pay 
for  244  pounds?  ^  Ans.  $30,50. 

This  is  worked  out,  in  the  book  from  which  we  ex- 
tracted it,  in  the  longest  and  most  formal  manner,  thus 
teaching  bad  habits  at  first,  which  are  difficult  to  eradicate ; 

and  until  such  are  eradicated,  it  is  impossible  to  be  skillful 
— 


126 


ARITHMETIC. 


As  another  specimen  of  our  ordinary  school  opera- 
tions, we  cut  the  following  problem  and  solution  from  a 
modern  popular  aritlimetic : 

"  If  3  hundred  weight  of  sugar  cost  £9  2  ^.  0  d.,  what 
will  4  hundred  weight  3  quarters  26  pounds  cost,  at  the 
same  rate  ? 

cwt.  dqr.  26/6. 

£  9 
20 


3  cwt.  4 

4  4 


2  s.  Od. 


12 


19 


4X7  = 


("7  7 

I     84       133 


182  s. 
12 


336/6.558/6. 


"  We  first  reduce 
the  first  and  second 
terms  to  pounds,  then 
the  3d  term  to  pence. 
The  answer  comes 
out  in  pence,  and  is 
afterwards  reduced  to 
pounds,  shillings  and 
pence. 


2184  (/, 
558 


2184 


^ns. 


17472 
10920 
10920 


336)1218672(3627  d. 
1008 
12)3627 


2106 
2016 


20)30  s.  3d. 


907                ^15  -  2s. 
672 


2352 
2352 


A7is.£l5  2  s.  3d.'* 

One  object  of  this  work  is  to  teach,  if  possible,  that 
kind  of  tact  possessed  by  business  men,  when  they  use 
thejr  judgments,  and  quickly  arrive  at  results.  The 
above  operation  is  strictly  legitimate,  and  agreeable  io 
rule;  but  nevertheless,  it  is  such  as  a  business  man,  or  a 


PROPORTION. 


127 


skillful  arithmetician,  wouM  never  go  through.  The  first 
caiiiion  is  strictly  to  examine  numhers.  We  perceive 
that  4  hundred  weight  3  quarters  26  pounds,  is  2  pounds 
less  than  5  hundred  weight.  Now  let  us  compute  the 
cost  of  5  hundred  weight,  and  deduct  the  cost  of  2  pounds. 
cwt.  cwt.  £. 
Therefore,  as  3  :  5  ::  9,1  :  the  4th  term. 
As  there  is  a  tenth  in  the  3d  term,  multiply  the  first  and 
third  by  10,  and  30 

Or  as 6 

For  the  2  pounds. 

lbs.     Ib'i.  shillings 
As  3X112  :  2  ::   182 
or,         168  :   1   ::   182 
or,  84  :   1  ::     91 

We  give  another  specimen  from  another  book. 

"  If  8  bushels  2  pecks  cost  $4,25,  how  many  bushels 
can  I  purchase  with  $38,25  ?  Jins.  76  bu.  2  pk. 

$.cts.  $.  cts.         bu.  pk. 

4,25      :      38,25     ::     8     2 
34  4 


5     ::     91 

:     the  4th  term. 

£.    s.     d. 

1      ::     91 

:     V=15     3     4 

£.     s.     d. 

'S. 

From  15      3     4 

take              1     1 

9  X  1     1 

Ans.  15      2     3 

1.5300 
11475 


34  pk. 


425)130050(306  pk. 
1275 

pk. 

2550  4)306 

2550      

76  bu. 

$.         $. 
We  operate  thus :  41  :  38^ 

or, 1     :  38'- 


As  two  denominations 
occur  in  the  third  term,  it 
is  reduced  to  the  less ; 
hence  the  result  is  pecks, 
which  must  be  reduced 
to  bushels. 


pk." 
bu. 
::  8^ 
:  2 


:  Ans. 

:  76^-  bu.  .9ns. 


(Art.  80.)  There  are  some  problems,  where  more  ap- 
parendy  requires  less,  and  at  first  view  demands  another 
rule  (jf  statement  from  that  given  in  Art.  79. 

Example.  If  6  men  build  a  wall  in  20  days,  how  long 


128  ARITH3IETIC. 


will  it  require  30  men  to  do  the  same?  Here  it  is  evi- 
dent that  more  men  will  do  the  work  in  less  time ;  and 
more  requires  less,  apparently.  By  the  same  nature  of 
things,  more  money  will  require  less  time  to  produce  the 
same  interest.  But  this  is  all  apparent;  there  is  no 
such  thing  in  nature  as  more  in  reality,  more  cause,  more 
force,  producing  less  in  effect 

Lithe  last  example,  the  mere  e^ris^ence  of  men  can- 
not be  compared  to  wall,  or  to  any  thing  they  may  do. 
Days'*  7vork  may  be  compared  to  wall,  or  to  any  thing 
accomplished  by  labor.  Money,  of  itself,  cannot  pro- 
duce interest ;  it  must  be  compounded  with  time ;  and 
all  such  problems  are  really  problems  in  compound  pro- 
portion, and  only  appear  to  be  in  simple  proportion,  by 
one  term  in  each  branch  of  the  question  being  the  same. 
(See  Art.  84.)     If  we  should  state  the  question,  thus  : 

men.   men.     days. 
As ....  6    :     30    ::    20    :     days  required,  as  a  care- 
less  or  unphilosophical   operator   might,  it  would    em- 
body no  sense,  as  it  would  imply  that  men  compare  with 
men,  as  days  compare  with  days,  without  any  reference 
to  what  may  have  been  done.     But  if  we  multiply  6  men 
by  the  number  of  days  they  worked,  (20,)  we  shall  have 
120  days'  work,  which  built  the  wall ,  but  when  this  120 
days'  work  was  performed  by  30  men,  it  w^ould  require 
them  4  days.      We  can  state  all  questions  by  direct  pro- 
portion^ provided  ive  use  philosophical  terms.      The 
question  under  consideration  should  be  stated  thus  : 
days^  ivork.      days^  work.   wall,     ivall. 
6X20       :       30X[]     ::      1      :      1. 
Here   is  a  proportion  wanting  a  factor  in  the  second 
term,  as  in  Art.  77,  and  can  be  found  in  the  same  way  as 
there  explained. 

Example  2.  If  a  man  perform  a  journey  in  6  days, 
when  the  days  are  8  hours  long,  how  many  d-iys  will  he 
require  to  perform  the  same  journey  when  the  days  are 
12  hours  long?  Here  the  journey  evidently  refers,  or  is 
measured,  by  the  number  of  hours  taken  to  perform  it; 
but  the  hours  in  both  branches  of  the  question,  is  the 
compound  of  days  and  hours.     The  statement  is  thus : 


PROPORTION.  129 


hoin's.     hours.      dis.   (lis. 

6X8  :  []X12  ::   1   :    1 

Solved  by  Art.  77. ^ns.  4. 

If  the  distances  had  been  diflerent,  not  exactly  the 
same  journey,  the  compound  nature  of  the  question 
would  have  been  apparent  at  once. 

(Art.  81.)  In  stating  a  question,  the  mind  should  rest 
upon  tilings  and  denominations,  and  not  on  their  numeri- 
cal values.  But  after  the  question  is  stated,  we  should  then 
look  on  the  terms  as  abstract  numbers,  and  nothing  else; 
and,  as  the  first  is  a  divisor,  and  (he  other  two,  factors 
in  a  dividend,  we  may  divide  the  first  by  any  number 
that  will  divide  it,  and  either  one  of  the  others,  without  a 
remainder,  (Art.  24.) 

Example  1.  What  will  26  yards  of  cloth  come  to,  if 
•$6,90  are  paid  for  13  ells  French  ?  To  compare  ells  with 
yards,  we  must  reduce  both  to  quarters;  but  in  many 
such  reductions,  the  multiplications  or  divisions  may  be 
only  formal,  not  reed.  Observe  the  following: 
(jr.  qr.  cts. 

6X13  :  26X4  ::  690 
Divide  1st  and  2d 
by  13, and 

or, 3  :         4      : :  690 

or, 1   :        4      ::  230 

As  we  have  before  observed,  this  is  the  same  as  can- 
(^eling,  (Art.  24 ;)  but  in  proportion,  this  appears  to  be 
most  convenient. 

Example  2.  If  15^  bushels  of  clover  cost  $1561;, 
what  quantity  can  be  bought  for  $95 '^^^  ? 

.^ns.  9,575  bush. 
Stateinent. 
156,25  :  95,75  ::   15,625  bu.  e^^^s. 
or,     10  :  95,75  ::      1     :  9,575  bu.  .^ns. 

^^'>-  <^      MISCELLANEOUS    EXAMPLES. 

1.  What  will  20  cords  of  wood  cost,  if  48  cords  are 
worth  120  dollars?  ./Ins.  $50. 

2.  If  20  cords  are  worth  50  dollars,  what  will  48  cords 
cost?  .^ns.  $120. 


2X4  ::  690 


Jins. 

Ans. 

Ans. 
$9,20  A. 


130  ARITHMETIC. 


3.  If  120  dollars  will  purchase  48  cords  of  wood,  how 
iPiany  cords  can  be  bough. t  for  50  dollars  ?       A)Vi.  20. 

4.  Thirty-six  gallons  of  wine  were  purchased  for  108 
dollars,  what  should  be  given  for  4  gallons  ?  Ans.  $12. 

5.  If  7  men  can  do  a  certain  piece  of  work  in  12  days, 
how  long  will  it  take  15  men  to  do  one-half  of  it?  (See 
Art.  80.)  Ans.  2i  days. 

6.  If  8  yards  of  cloth  cost  3^  dollars,  how  many  yards 
can  be  bought  for  50  dollars  ?  Ans,  1 14|  yd. 

7.  A  garrison  has  provision  for  8  months,  at  the  rate 
of  15  ounces  per  day ;  what  must  be  each  man's  allow- 
ance, that  the  same  provision  may  last  them  12  months  ? 

Ans.  10  ounces. 
By  Art.  83,  this  problem  is  solved  thus : 


oz.  oz.        provis.  pi'ovis. 

8X15  :   12X[]   ::   1      :      1.      Hence,  12 


,  _  ^  =Ans. 


Keep  the  same  Art.  in  mind  in  solving  the  two  follow- 
ing: 

8.  When  a  quarter  of  wheat  affords  60  ten-penny 
loaves,  how  many  eight-penny  loaves  may  be  made  from 
the  same  ?  Ans.  75. 

9.  If  10  dollars'  worth  of  provision  serve  7  men  4 
days,  how  many  days  will  the  same  provision  serve  9 
men  ?  Ans.  3^  days. 

IQ.  If  15  degrees  of  longitude  cause  a  difference  in 
time  of  1  hour,  what  will  be  the  hour  at  Washington, 
when  it  is  12  o'clock  at  a  place  situated  36°  45'  east  of  it? 

Ans.  9  o'clock  33  min. 

11.  If  a  man's  yearfy  income  b#  $2000,  and  his  aver- 
age expenditures  be  S25,87^  per  week,  how  much  can 
he  lay  up  in  6  years,  including  one  leap-year  ? 

Ans.  ^4057,621. 

12.  If  17  men  do  a  piece  of  work  in  20  days,  in  what 
time  will  20  men  do  double  the  work  ?  Ans.  34  days. 

13.  If  a  staff  3  feet  8  inches  long,  cast  a  shadow  1  foot 
6  iuciies,  what  is  the  height  of  a  steeple  that  casts  a  shad- 
ow 75  feet  at  the  same  time  ?  Ans.  183'^. 

14.^4f  I  of  an  ell  English  cost  \  of  a  pound,  what  will 
12|  yards  cost?  Ans.  £5  lis.  l\  d. 


PROPORTION.  131 


15.il If  12!,  hundred  weight  of  iron  cost  $42] ,  what  will 
48;'5-  liundred  weight  cost?  ./Ins,  $163,50. 

16.  What  quantity  of  Hning,  f  yard  wide,  will  it  re- 
quire to  line  9|.  yards  of  clotli,  1-^  yard  wide  ? 

.r]ns.  15^  yards. 

17.  If  12  gallons  cost  $30,  what  will  63  gallons  cost  ? 

JIns.  $157,50. 

18.  How  many  yards  of  clolli,  3  quarters  wide,  are 
equal  to  30  yards  5^ quarters  wide  ?       .^ns.  50  yards. 

lO.'How'many  yards  of  paper,  |  of  a  yard  wide,  will 
be  sufficient  to  "hang  a  room  20  yards  square,  and  3^ 
yards  high  ?  ./^ns.  400  yards. 

20.  If  28  dollars  are  paid  for  15  gallons  of  wine,  what 
is  the  price  per  pint  ?  *^ns.  23}  cents. 

21.  If  240  bushels  of  wheat  are  purchased  at  the  rate 
of  $22V  for  18  bushels,  and  sold  at  the  rate  of  $33|  for 
22  V  bushels,  what  is  the  profit  of  the  whole  ? 

Ms.  60  dollars. 

22.  If  $100  gain  $7  interest  in  a  year,  what  will  $49,- 
75  gain  in  the  same  time  ?  ^n-s.  $3,48 'j. 

23.  If  a  hogshead  of  brandy  cost  ^61 6  16  shiUings,  what 
Mdll  12  gallons  cost  ?  ^ns.  £3  4  s. 

24.  If  27  yards  of  Irish  linen  cost  £5  12  shillings  6 
pence,. how  many  ells  English  can  be  bought  for  £100  ? 

Ms.  384. 

25.  If  300  cwt.  be  carried  660  miles,  for  4  dollars,  how 
many  hundred  can  be  carried  60  miles  for  the  same  mo- 
ney? ^ns.  3300. 

i^26.  What  quantity  of  water  added  to  31-^-  gallons  of 
whisky,  which  cost  $13,50,  would  enable  the  purchaser 
to  sell  it  for  40  cents  per  gallon,  at  a  profit  of  10  cents  a 
gallon  on  his  purchase?  Ms.  10^^  gallons. 

27.  If  it  require  90  yards  of  carpeting,  J  wide,  to  car- 
pet a  floor,  how  many  yards,  11  wide,  would  be  suffi- 
cient ?  *^^i^'  60  yards. 

28.  If  9 1  yards  of  linen  cost  2  pounds  12  shillings, 
what  will  32 1'  yards  cost  in  dollars,  at  4  shillings  6  pence, 
each?  .^775.  $382. 

29.  At  $2,75  for  14  pounds  of  sugar,  what  would  be 
the  price  of  I  hundred  weight?  Ms.  $22. 

/    30.  If  A  can  Aow  ^,  B  i,  and  C  j  of  an  acre  of  grass 

I ; — 


132  ARITHMETIC. 


in  an  hour,  in  how  many  hours  can  ihey  together  mow 
'S\  acres  ?  ^ns.  6  h.  30  m.  30;,  sec. 

J  ^  3  Ip*  IX  A^,  and  B  together  can  do  a  piece  of  work  in  7 
"X.  1  9^ysv^t;id;^  alone  in  12  days,  in  how  many  days  can  A 
alone  do  f  of  it?  |  ^9ns.  1^  days. 

32.  If  5  hundred  weight  3  quarters  14  ppunds  of  su- 
gar cost  6  pounds  1  shilling  8  pence,  how  many  dollars, 
at  6  shillings  each,  will  35  hundred  weight  28  pounds 
cost?  ^.;  c^n*.  $121f. 

N.  B.  See  example  2,  Art.^TQ. 

33.  A  merchant  bought  of  a  farmer  120  bushels  of 
wheat  at  75  cents  per  bushel,  and  sold  it  at  20  per  cent, 
advanced  price  ;  what  did  he  receive  for  the  whole,  and 
how  much  did  he  gain  ? 

Statement,  100  :  120  :  :  75  :  to  the  merchant's  price, 
90  cents  ;  whole  gain,  $18.  .\,  | 

34.  A  merchant  bought  sugar  at  $4,75  per  100  pounds; 
how  much  shall  he  sell  it  per  pound  to  gain  25 'per  cent.? 

^<^ns.  6  cents,  very  nearly. 

35.  ^Bought  tea  at  62^  cents  per  pound,  but  by  a  fall 
-ra  the  market,  I  am  willing  to  sell  it  at  10  per  cent,  loss ; 
'what  shall  be  the  price  of  it  ?  Jins.  56^  cents. 

N.  B.  The  three  preceding  problems  could  properly 
be  put  under  the  technical  head  of  loss  and  gain  ;  but  it 
is  as  propp7'  that  they  should  be  under  proportion. 

36.  A  pole,  whose  height  is  25  feet,  at  noon  casts  a 
shadow  to  the  distance  of  33  feet  10  inches  :  what  is  the 
breadth  of  a  river  which  runs  due  East,  at  the  bottom  of 
a  tower  250  feet  high,  whose  shadow  extends  just  to  the 
opposite  edge  of  the  water  ?  ^^ns.  338  ft.  4  in. 

37.  A  plane  of  a  certain  extent  having  supplied  a  body 
of  3000  horses  with  forage  for  18  days ;  how  long  would 
it  have  supplied  2000  horses?  .^liis.  27  days. 

38.  If  a  carriage- wheel  in  turning  twice  round,  advance 
33  feet  10  inches  ;  how  far  would  it  go  in  turning  round 
63360  times  ?  .^ns.  203  miles. 

39.  Sound  flies  at  the  rate  of  1142  feet  in  1  second  of 
time:  I  heard  the  report  of  a  gun  1  minute  and  3  seconds 
after  I  saw  the  flash ;  how  far  distant  was  it  ? 


PROPORTION.  133 


40.  If  a  carter  haul  100  bushels  of  coal  at.  every  3  loads, 
how  many  days  will  it  require  for  him  to  load  a  boat  with 
3600  busiiels,  suppose  he  hauls  9  loads  a  day  ? 

^ns.  12  days. 

41.  If  the  moon  move  at  the  average,  rate  of  13°  10' 
35"  in  one  day,  how  lonjr  will  it  require  to  perform  a  re- 
volution ?        '  Ans.  27  d.  7  h.  43  m.  nearly. 

42.  The  mean  motion  of  the  moon  being-,  as  stated  in 
the  last  problem,  13°  10'  35"  per  day,  the  mean  apparent 
motion  of  the  sun  in  the  same  time  59'  8"33,  wliat  time 
will  be  required  for  the  moon  to  gain  a  revolution  on  the 
sun  ?  Ans.  29  d.  12  h.  44  m.  3  s. 

Questions. 

What  is  meant  by  comparison  of  numbers?  Ans.  An 
examination  of  their  relative  magnitudes  or  values. 

When  such  magnitude  is  expressed,  what  is  the  ex- 
pression called  ?  Ans.  Proportion,  or  ratio.  It  is  called 
ratio,  when  one  number  is  expressed  by  unity. 

Is  the  ratio  of  two  numbers  the  same  as  the  quotient 
resulting  from  the  division  of  one  of  them  by  the  other  ? 

What  name  is  given  to  the  first,  and  what  to  the  second 
term  of  a  couplet  ? 

When  two  pairs  of  numbers  have  the  same  ratio,  what 
do  they  constitute? 

What  name  is  given  to  the  two  outer  terms,  and  what 
to  the  two  middle  terms  of  a  proportion  ? 

What  relation  is  there  between  the  product  of  the 
means,  and  the  product  of  the  extremes  of  a  proportion  ? 

Why  is  the  product  of  the  means  equal  to  the  product 
of  the  extremes  ? 

AVhat  rule  is  founded  on  this  equality  ? 

Explain  the  common  method  of  stating  in  the  Single 
Rule  of  Three. 

How  do  v.e  proceed  after  a  question  is  stated  ? 

Why  must  the  two  like  terms  be  brought  to  the  same 
denomination  ? 

When  we  have  the  1st,  2d,  and  4th  terms  of  a  propor- 
tion, how  do  we  find  the  third  ? 

When  a  factor  of  any  one  term  is  wanting,  how  is  it 
found  ? 

M 


134  ARITHMETIC. 


COMPOUND  PROPORTION: 

OR, 

THE  DOUBLE  RULE  OF  THREE. 

(Art.  82.)  The  following  rule  is  common  to  all  modern 
authors,  though  they  may  make  slight  variations  in  ex- 
pressing it.  We  give  it  in  the  words  of  Mr.  G.  "Wilson, 
v/ith  an  example  and  its  accompanied  explanation,  as  ex- 
pressed in  great  clearness  and  brevity. 

Compound  Proportion  is  compounded  of  two  or  more 
ranks  of  proportionals,  five,  seven,  nine,  &;c.,  terms  be- 
ing given  to  find  a  sixth,  eighth,  &c. 

Rule.  Work  by  two  or  more  statements  in  Simple 
Proportion ;  or,  set  that  term,  which  is  like  the  term 
sought,  in  the  third  place,  and  consider  each  pair  of 
similar  terms  and  this  third  one,  as  the  terms  of  a  state- 
ment  in  Simple  Proportion,  and  set  them  severally,  in 
the  first  and  second  places,  agreeably  to  the  directions 
under  that  rule. 

TVhen  the  question  is  thus  stated,  reduce  the  similar 
terms  to  the  like  denominations,  and  then  multiply  all 
the  terms  in  the  second  and  third  places  together,  arid 
divide  the  product  by  the  product  of  those  in  the  first 
place;  the  quotient  will  be  the  answer,  or  term  sought. 

The  method  of  statement  will  best  be  illustrated  by  an 
example. 

1.  If  8  men  can  build  a  wall  20  feet  long,  6  feet  high, 
and  4  feet  thick,  in  i2  days,  in  what  time  v/ill  24  men 
build  one  200  feet  long,  8  feet  high,  and  6  feet  thick  ? 

Here  the  answer  sought     men  24  :       8"^  , 

is  days.      We,   therefore,     length,      20  :  200  !         ^^ 
place  12  days  for  the  third     ^    •  '  ■ 
term,  and   comparing    the 
number  of  men,  it  is  evi- 
dent that  24  men  will  re- 
quire less  time  to  do  the  same  piece  of  work  than  8  men. 
We  therefore  place  the  less  of  the  two  numbers   last. 


men          24  . 

8^ 

length,      20  : 

200  ! 

height,        6  : 

thickness,  4  : 

6J 

COMPOUND     PROPORTION.  135 


We  next  compare  the  length  of  the  two  walls,  and  see 
that  one  200  leet  long  will  require  more  clays  than  one 
20  feet  long,  and  place  the  longest  last.  For  the  same 
reason,  we  arrange  the  heights  of  the  two  walls  in  the 
same  order;  and  so  also  of  the  thickness.  The  con- 
tinued product  of  all  the  middle  terms  and  the  third  term, 
divided  by  the  product  of  all  the  first  terms,  gives  the  an- 
swer 90  days. 

But  the  labor  of  multiplication  (as  in  simple  propor- 
tion) may  be  shortened  by  reducing  any  of  the  first  terms, 
and  any  of  the  other  terms  proportionally  by  division. 
Or  we  may  draw  a  perpendicular  line,  and  cancel  as  in 
Art.  24;  the  full  multiplications  and  divisions  should 
never  be  accomplished,  unless  the  numbers  are  prime  to 
each  other. 

The  preceding  method  is  very  good,  as  a  7nechanicol 
contrivance^  but,  as  a  philosophical  expression,  it  will  not 
bear  criticism.  We  prefer  the  following,  by  reason  of  its 
greater  simplicity  and  more  extensive  application. 

(Art.  83.)  It  is  an  axiom  in  philosophy,  that  equal 
causes  produce  equal  effects;  and  that  effects  are  always 
proportionate  to  their  causes. 

Now,  causes  and  effects  that  admit  of  computation, 
that  is,  involve  the  idea  of  quantity,  may  be  represented 
by  numbers,  which  will  have  the  same  relation  to  each 
other  as  the  things  they  represent. 

Keeping  these  premises  in  view,  then,  we  have  an  uni- 
versal rule,  applicable  to  all  cases  which  can  arise  un- 
der Proportion,  simple  or  compound,  direct  or  inverse, 
namely : 

Rule.  As  any  given  cause  is  to  its  effect,  so  is  any 
required  cause  (of  the  same  kind)  to  its  effect;  or,  so 
is  another  given  cause,  of  the  same  kind,  to  its  required 
effect. 

The  only  difficulty  the  pupil  can  experience  in  this 
system  of  proportion,  is  readily  to  determine  what  is 
cause,  and  what  is  effect.  But  this  difilculty  is  soon 
overcome,  when  we  consider  that  all  action,  of  whatso- 
ever nature,  must  be  cause — and  wdiatever  is  accomplish- 
ed by  that  action,  or  follows  such  action,  must  be  eifect. 


136  ARITHMETIC. 


EXAMPLES. 


1.  If  16  horses,  in  50  days,  consume  128  bushels  of 
oats,  how  many  bushels  will  5  horses  consume  in  90 
days  ?  Ans.  TZ. 

Here  it  is  evident,  that  the  consumption  of  oats  spoken 
of,  in  both  the  supposition  and  demand,  are  the  true  ef- 
fects ;  and  the  action  of  the  horses,  multiplied  by  the 
days,  must  express  the  amount  of  cause.  We  shall, 
therefore,  state  it  thus : 


Cause. 

Effect. 

Cause. 

Effect. 

16 

:       128 

::         5        : 

n 

50 

90 

We  write  the  factors,  one  under  another,  as  above ; 
their  multiplication  is  understood,  but  rarely  or  never  ac- 
tually  accomplished.  The  second  effect  is  an  unknown 
term,  or  answer,  required — a  bracket,  or  blank,  or  point, 
is  left  to  represent  it.  When  found,  the  four  terms  above 
would  be  2.  perfect  geometrical  proportion,  and  the  pro- 
duct of  the  extremes  equal  to  the  product  of  the  means. 
In  this  example,  the  product  of  the  means  is  perfect ; 
which  product,  divided  by  the  factors  in  the  extremes, 
will  give  what  is  wanting  in  the  extremes,  namely — the 
answer. 

Therefore,  agreeably  to  our  mode  of  performing  mul- 
tiplication and  division,  we  draw  a  line  thus,  and  cancel 
down : 


16 
50 


128 
5 
90 


2.  If  $480,  in  30  months,  produce  $84  interest,  what 
capital,  in  15  months,  will  produce  $21  ? 

Now,  capital  will  not  produce  interest  without  time; 
and,  whatever  be  the  rate  per  cent.,  the  same  capital  in  a 
double  time  will  produce  a  double  interest.     Therefore, 

Cause.  Effect.  Cause.  Effect. 

480        :        84         :  :  [  ]         ;         21 

30  15 


COMPOUND    PROPORTION.  137 


Here  one  element  of  the  second  co.use  is  wanting;  that 
is,  tlie  answer  to  the  question. 

In  this  case,  the  extremes  ar&  complete  ;  we  will  there- 
fore, divide  the  product  of  the  extremes  by  the  factors  in 
the  mean,  and  tlie  quotient  will  give  the  definite  factor  ^ 
or  answer,  namely — $240. 

3.  If  7  men,  in  12  days,  dig  a  ditch  60  feet  long,  8 
feet  wide,  and  6  feet  deep,  in  how  many  days  can  21 
men  dig  a  ditch  80  teet  long,  3  feet  wide,  and  8  feet  deep  ? 

It  is  almost  too  plain  for  comment,  that  7  men,  multi- 
plied by  12  days,  must  be  the  first  cause,  and  the  con- 
tents of  the  ditch  they  dig,  the  effect.     Therefore, 

Cause.         Effect.  Cause.         Effect. 

7        :        60  :  :  21  :        80 

12  8  []  3 

6  8 

Here,  as  in  the  preceding  example,  one  of  the  elements 
of  the  second  cause  is  wanting ;  or,  rather  say  a  factor 
in  the  means  of  a  perfect  proportion,  and  can  be  found 
as  above.  Jins.  2|  days. 

4.  If  6  men  build  a  wall  in  12  days,  how  long  will  it 
require  20  men  to  build  it?  Ans.  3|  days. 

(Art.  84.)  Questions  of  this  kind  are  usually  classed 
under  the  single  rule  of  three  inverse  ;  they  do,  in  fact, 
however,  belong  to  compound  proportion:  but,  as  one 
of  the  terms  is  the  same  in  the  supposition  as  in  the  de- 
mand, it  may  be  omitted.  That  term,  in  the  present  ex- 
ample, is,  one  wall.  If  we  make  the  number  different 
in  the  two  brandies  of  the  question,  or  connect  any  con- 
ditions with  it,  such  as  lengths,  breadths,  &c.,  it  at  once 
falls  under  compound  proportion  of  necessity,  and  may 
be  stated  thus : 

Cause.         Effect.  Cause.         Effect. 

6:1  :  :  20  1 

12  [] 

5.  If  4  men,  in  2|  days,  mow  6|  acres  of  grass,  by 
working  8{  hours  a  "day,  how  many  acres  will  15  men 
mow  in  3|  days,  by  working  9  hours  a  day? 

Ans.  40 yy  acres. 

ITT- 


138  ARITHMETIC. 


Cause.         Eject.  Cause.         Effect. 

4         :         6f  ::  15         :        [] 

^  3J 

81  9 

Let  the  pupil  observe,  that  when  a  correct  statement 
is  made,  there  will  be  the  same  number  of  elements,  or 
factors,  under  the  same  letters,  as  in  the  above  example. 
Under  each  cause,  we  have  men,  days,  and  hours,  to 
be  multiplied  together.  When  there  are  fractions  in  any 
of  the  terms,  their  denominators  are  to  be  placed  over 
the  line  from  where  the  term  belongs;  mixed  numbers 
being  previously  reduced  to  improper  fractions. 

6.  If  12  ounces  of  wool  make  H  yards  of  cloth,  |-  of 
a  yard  wide,  how  many  yards,  l\  wide,  will  16  pounds 
of  wool  make  ?  Ans.  22 1  yards. 

With  this  example,  some  might  hesitate  as  to  arranging 
it  under  cause  and  effect,  as  the  actors,  those  who  made 
the  cloth,  whether  many  or  few,  have  nothing  to  do  with 
the  question.  But  we  take  the  phrase  of  the  example, 
and  say,  The  wool  makes  the  cloth. 

Cause.         Effect.  Cause.       Effect. 

12        :         H  ::  16>  [] 

i  16^     •  H 

7.  If  the  transportation  of  124  hundred  weight,  206 
miles,  cost  $25,75,  how  far,  at  the  same  rate,  may  3  tons 
and  3  quarters  be  carried  for  $243  ?  Ans.  402f  miles. 

In  this  example,  it  is  indifferent  which  we  take  for  the 
cause,  and  which  for  the  effect.  We  may  say,  that  the 
money,  825,75,  is  the  cause  of  having  the  weight  car- 
ried ;  or,  we  may  say,  that  carrying  the  weight  is  the 
cause  of  purchasing  the  money.  There  are  many  ques- 
tions where  it  is  indifferent  which  we  take  for  cause,  and 
which  for  eflect.     The  above  example  may  be  stated  thus : 

Cause.         Effect. 
: :       243     :     60|  cwt. 
[] 

Cause.         Eff'ect. 
60J       :       243 
206'  ^  [] 


Cause. 

Effect. 

25^- 

:     124  cwt. 

206  miles 

Or  thus : 

Cause. 

Effect. 

'       12± 

:       25| 

COMPOUND    PROPORTION.  139 


Or  thus : 

Cause. 

12i 
206 

Cause. 
:       60?- 

[] 

Or  thus : 

Effect. 
243 

Effect. 
:       251 

Effect.         Effect. 
251       :       243 


Cause.         Cause. 
:         60|       :       121 
[]  206 

These  changes  show,  conclusively,  that  this  method  of 
statement  is  strictly  scientific  and  philosophical ;  and,  in 
all  these  different  arrangements  of  the  terms,  the  scnne 
terms  are  multiplied  together. 

The  most  that  can  be  said  for  the  common  mode  of 
statement,  in  the  Double  Rule  of  Three,  is,  that  ihe pro- 
ducts, when  the  terms  are  midtiplied  out,  are  propor- 
tional. But  the  first  and  second  terms,  taken  as  a  whole, 
express  no  particular  idea  or  thing ;  whereas,  in  this  mode 
of  statement,  the  thing — the  philosophical  idea — is  the 
only  sure  guide.  Nor  is  this  all ;  it  is  very  extensive, 
and  easy  in  its  application ;  it  will  cover  every  case  that 
can  arise  under  interest — to  find  time,  rate  per  cent.,  &c., 
and  thus  do  away,  or  suspend,  five  or  six  special  rules, 
which  encumber  every  arithmetic.  We  give  a  few  ex- 
amples, to  apply  this  rule. 

8.  What  is  the  interest  of  240  dollars  for  3-2  years,  at 
6  per  cent.? 

Cause.         Effect.  Cause.         Effect. 

100        :        6  ::  240       :       [] 

1  3t 

To  obtain  the  answer  from  this  statement,  we  perceive 
that  we  must  multiply  the  means  together — i.  e.  240  X  6, 
the  rate,  and  that  by  the  3i  years,  the  time — and  divide 
by  100;  and  this  is  the  common  ride. 

9.  The  interest  of  a  certain  sum  of  money,  at  6  per 
cent.,  for  15  months,  was  60  dollars  ;  what  was  the  sum  ? 

^ns.  $800. 
Cause.         Effect,  Cause.         Effect. 

100        :        6  ::  [  ]       :       60 

12  15 


140  ARITHMETIC. 


10.  If  800  dollars,  in  15  months,  should  gain  60  dol- 
lars, what  would  be  the  rate  per  cent.?  Ans.  6. 

Cause.         Effect.  Cause.         Effect. 

800       :        60  :  :  100       :[  ] 

15  12 

11.  Eight  hundred  dollars  was  put  out  at  interest,  at  6 
per  cent.,  and  the  interest  received  was  60  dollars  4  how 
long  was  it  out  ?  Ans.  15  months. 

Cause.         Effect.  Cause.         Effect. 

100        :        6  ::  800       :         60 

12  [] 

12.  If  12  men,  working  9  hours  a  day,  for  15|  days, 
were  able  to  execute  f  of  a  job,  how  many  men  may  be 
vnthdravm,  and  the  residue  be  finished  in  15  days  more, 
if  the  laborers  are  employed  only  7  hours  a  day  ? 

Ans.  4  men. 

13.  The  amount  of  a  note,  on  interest  for  2  years  and 
6  months,  at  6  per  cent.,  is  690  dollars  ;  required  the 
principal.  Ans.  S600. 

14.  What  is  the  interest  of  1248  dollars,  for  16  days? 

Ans.  $3,28. 

Cause.         Effect.  Cause.         Effect. 

100        :        6         :;  1248      :       [] 

12?   ,  16 

30^  days. 

This  can  be  canceled  down,  and  made  very  brief. 

15.  What  is  the  interest  of  1200  dollars,  for  15  days, 
at  6  per  cent.?  Ans.  $3,00. 

16.  What  is  the  interest  for  160  dollars,  for  36  days, 
at  7  per  cent.?  Stated  by  cause  and  effect.  Ans.  $1,12. 

Cause.         Effect.  Cause.         Effect. 

100        :        7  ::  160       ;       [] 

12  1.2 

17.  The  interest  of  a  certain  sum,  for  36  days,  is 
$1,12 — the  rate  per  cent,  is  7  ;  what  is  the  sum  ? 

Ans.  $160, 

18.  The  interest  of  160  dollars  for  36  days,  is  $1,12  ; 
what  was  the  rate  per  cent.?  Ans.  7. 


COMPOUND    PROPORTION.  141 


19.  The  interest  of  160  dollars,  at  7  per  cent.,  was 
$1,12;  what  was  the  tune?  JIns.  36  days. 

20.  In  what  time  will  any  sum  double  itself  at  6  per 
cent.?     At  any  per  cent.? 

JIns.  At  6  per  cent.,  in  16|  years;  at  any  per  cent., 
divide  100  by  the  per  cent. 

21.  If  2^  yards  of  cloth,  1|  wide,  cost  3  dollars  37^ 
cents,  how  much  will  36",  yards  cost,  1  i  yards  wide  ? 

^Ans.  $52,79. 

22.  If  1  men  spend  f  of  f  of  f  of  Vi  of  £30,  in  -fy 
of  ^-  of  14  of  y  of  9  days,  how  many  dollars,  at  6  shil- 
lings each,  will  21  men  spend  in  |  of  \^  of  ^  of  yS  of  45 
days?  Ans.  $630. 

23.  I  lend  a  friend  200  dollars  for  6  months  ;  how 
long  ought  he  to  lend  me  1000  dollars,  to  requite  the  fa- 
vor— allowing  30  days  to  a  month?        Ans.  36  days. 

24.  If  1000  men,  besieged  in  a  town,  with  provisions 
for  5  weeks,  allowing  each  man  16  ounces  per  day,  be 
reinforced  with  500  men  more — and  supposing  that  they 
cannot  be  relieved  until  the  end  of  eight  weeks — how 
many  ounces  a  day  must  each  man  have,  that  the  provis- 
ions may  last  them  that  time?^  Ans.^S\  ounces. 

25.  If  a  foot-man,  in  12^days,  traveling  ^lours  a  day, 
perform  a  journey  of  240'miles,  in  how  many  days  will 
he  perform  one  of  720^Tiiles,  if  he  travel  8  hours  a  day? 

Ans.  '21  days. 
^^  26.  If  20  men,  in  12  days,  working  5  hours  a  day, 
can  perform  a  piece  of  work,  how  many  hours  a  day 
must  15  men  work,  in  order  to  perform  3^-  times  as  much 
work  in  30  days  1  Ans.  8-^  hours. 

27.  In  what  time  will  $627,50,  loaned  at  7  per  cent., 
pro<luce  as  much  interest  as  $2510,  at  3^  per  cent.,  will 
produce  in  1  year  and  8  months?  Ans.  3|-  years. 

28.  If  a  block  of  marble  2  feet  6  inches  long,  1  foot  9 
inches  broad,  and  1  foot  3  inches  thick,  weigh  9  hundred 
weight  2  quarters,  what  would  it  weigh,  if  each  of  its  di- 
mensions were  doubled?  Ans.  3  tons,  16cwt. 

29.  Eight  workmen,  laboring  7  hours  a  day  for  15 
days,  were  able  to  execute  \  of  a  job  on  which  they  were 
engaged ;  in  how  many  days  can  they  complete  the  res- 


142  ARITHMETIC. 


idiie.  by  working  9  hours  a  day,  if  4  workmen  are  added 
to  their  number?  Jlns.  15|  days. 

30.  How  many  men  will  reap  417,6  acres  in  12  days, 
if  5  men  reap  52,2  acres  in  6  days  ?       Ans.  20  men. 

31.  If  a  cellar  22,5  feet  long,  17,3  feet  wide,  and  10,25 
feet  deep,  be  dug  in  2,5  days,  by  6  men,  working  12,3 
hours  a  day,  how  many  days,  of  8,2  hours,  should  9  men 
take  to  dig  another,  measuring  45  feet  long,  34,6  wide, 
and  12,3  deep?  Ans.  12  days. 

32.  If  54  men,  in  24^  days,  working  12^  hours  each 
day,  can  build  a  fort,  in  how  many  days  will  75  men  do 
the  same,  when  they  work  but  10|  hours  each  day  1 

Ans.  21  days. 

33.  If  24  men,  in  189  days,  working  14  hours  each 
day,  dig  a  trench  33 1  yards  long,  3^-  deep,  and  5|  wide, 
how  many  hours  per  day  must  217  men  work,  to  dig  a 
trench  23i  yards  long,  21  deep,  and  3|  wide,  in  5^  days? 

Ans.  16  hours. 

34.  If  a  cistern,  16  feet  long,  7  broad,  and  15  deep, 
cost  36  dollars  72  cents,  how  much,  in  proportion,  would 
another  cistern  cost,  that  is  17^  feet  long,  10^  broad,  and 
16  deep?  Ans.  %M,2Q. 

35.  If  a  cistern  17^  feet  long,  10^  broad,  and  13  deep, 
hold  546  barrels,  how  many  barrels  will  that  cistern  hold, 
that  is  16  feet  long,  7  broad,  and  15  deep? 

Ans.  384  bbls. 
(Art.  85.)  With  the  exception  of  square  and  cube 
root,  and  a  few  obvious  principles  in  mensuration,  we 
have  now  explained  every  arithmetical  operation.  The 
four  groimd  rules,  with  reduction,  and  proportion,  in- 
clude all  the  principles  that  can  be  used  in  Practice,  In- 
terest, Barter,  Loss  and  Gain,  Exchange,  &c.  &c.;  hence 
when  we  come  to  these  subjects,  we  have  no  new  prin- 
ciples to  explain.  It  only  remains  to  give  clear  defini- 
tions under  each  head,  and  teach  how  to  apply  the  knowl- 
edge we  are  supposed  already  to  possess  ;  and  now  all 
the  judgment  and  tact  of  the  jpupil  will  be  called  into  full 
exercise. 


PRACTICE. 


143 


PRACTICE. 

(Art.  86.)  Practice  is  literally  what  the  term  implies: 
short  and  expeditious  methods,  practiced  bv  merchants 
and  business  men,  to  find  tlie  amount  of  bills,  and  the 
aggregate  cost  of  articles  bougJal  or  sold.  The  principle 
involved  in  all  such  calculations,  is  proportion ;  but  the 
price  of  the  unit  is  generally  given,  which  renders  a  i'or- 
mal  statement  unnecessary;  and  the  computations  ;ire 
usually  made  by  taking  the  number  of  times,  and  the 
aliquot  parts  of  the  unit. 

TABLE  OF  ALIQUOT  (OR  EVEN)  PARTS. 


Cents. 

Parts 

of$l. 

0 

6 

Parts 
of  1  year. 

Parts  of  £1. 

Parts  of  1 

shilling-. 

Parts  of  1  cwt. 

50 

V 

i 

lOs.       =V 

6d.   =i 

561b.  =  !, 

33L 

4 

4 

3" 

6s.8d.=i 

4d.   =• 

28lb.  =  L 

25 

4 

3 

4 

5s.       =L 

3d.   =• 

161b.=l 

20 

T 

2 

6 

4s.       i=i^ 

2d.   =J 

14lb.  =  i 

12\ 

■r- 

1 

tV 

3s.4d.=i. 

lid.=i 

7lb.  =  J, 

6't 

Te 

or  i  of 

2s.6d.=i 

Id.  =yV 

5 

I 

20 

3  mo. 

ls.8d.=J^ 

(Art.  87.)  The  principles  of  computation  being  sup- 
posed to  be  well  fixed  in  the  mind,  2Lny  formal  rule  that 
would,  in  any  degree,  restrain  the  free  exercise  of  judg- 
ment, in  the  almost  endless  variety  of  cases,  would  prove 
an  nijury  rather  than  an  aid ;  therefore  we  shall  explain 
only  by  examples. 

Example  1.  What  will  64  bushels  of  oats  cost,  at  25 
cents  per  bushel  ?  Ans,  \  as  many  dollars  as  bushels, 
16  dollars. 

2.  What  will  47^  bushels  of  wheat  come  to,  at  87^, 
cents  ? 

At$l, $47,50 

iless, 5,94 


Jlns.  $41,56 


144  ARITHMETIC. 


3.  What  will  78  gallons  of  wine  cost,  at  1  dollar  62 J- 
cents  per  gallon  ? 

At  $1,    ...  • $78,00 

At  50  cts 39,00 

At  12^, 9,75 


Ans,  $126,75 
4.  AVhat  will  1  bushel  3  pecks  6  quarts  of  beans  cost, 
at  1  dollar  12 1  cents  per  bushel  ?     This  is  2  quarts  less 
than  2  bushels  :  hence,  2  bushels  cost $2,25 


of  1,12|,  less,  or  7  cents,  nearly, 7 


Ans.  $2,18 

5.  What  will  75  bushels  cost,  at  331  cents  a  bushel  ? 

3)75 

25  dollars,  Ans. 

6.  What  will  224  barrels  of  flour  come  to,  at  3  dollars 
43^  cents  per  barrel? 

224  at  $3, 672 

at  50  cts 112 


784 
-r'^  of  224,  less, 14 

Ans.  $770 

7.  What  will  462  yards  of  cloth  cost,  at  $1,1 2  V  per 
yard?  Ans.  $519,75. 

8.  What  will  185  yards  of  cloth  cost,  at  $1,20  per 
yard?  Ans.  $222. 

9.  What  will  150  yards  of  cloth  cost,  at  $1,3^  per 
yard?  Ans.  $196,87^-. 

10.  What  is  the  cost  of  144  pounds  of  rice,  at  3V  pence 
a  pound ?     This  had  better  be  done  by  canceling ;  3^  =  |. 

4^  12 

Thus, ^1       ' 

20  Ans.  £2  2  s. 

11.  What  will  45^  barrels  of  flour  come  to,  at  6  dol- 
lars 25  cents  a  barrel  ?    '  Ans.  $284,37|. 


PRACTICE.  145 

12.  What  will  18^  cords  of  wood  cost,  at  4  dollars  75 
cents  a  curd  I  "  Jlns.  $87,87  ^ 

13.  What  will  "3 J  tons  of  coal  cost,  at  4  dollars  25 
cents  per  ton  ?  Jins.  $15,93^. 

14.  Wliat  will  2  hundred  weight  2  quarters   14  pounds 
of  sugar  come  to,  at  $8,50  per  hundred  weiglit? 

Jim.  $22,31  V. 

15.  What  will  3  bushels  3  pecks  6  quarts  of  beans 
cost,  at  $1,60  per  bushel  ?     [Find  what  4  bushels  cost.] 

Am.  $G,30. 

(Art.  88.)  In  some  States,  particularly  in  New  Eng- 
land and  New  York,  the  prices  of  articles  arc  still  given 
in  sliillings  and  pence;  yet  they  never  render  a  bill  in 
})ound3,  sliillings  and  pence,  but  always  chaufj-e  it  to  dol- 
lars and  cents.  In  New  England,  as  there  are  6  shillings 
to  a  dollar,  6  yards  or  6  pounds  will  cost  as  many  dol- 
lars as  there  are  shillings  and  parts  of  a  shilling  to  one 
yard,  pound,  &.c. 

Excnnple  1.  At  5  shillings  6  pence  per  yard,  what  will 
6  yards  cost,  in  dollars  and  cents  ?  Jlns.  $5,50, 

N.  B.  The  5  shillings  are  called  5  dollars,  and  0  pence 
are  called  50  cents. 

Example  2.  At  7  shillings  3  pence.  New  Egland  cur- 
rency, per  yard,  what  will  18  yards  of  cloth  cost? 

6  yards  will  cost $7,25 

18  yards  is  3  times  as  much 3 

Am. $21,75 

Example  3.  What  will  15  bushels  of  wheat  cost,  at  5 
f  hillings  9  pence  (N.  E.)  per  bushel  ? 

6  bushels  cost $5,75 

6  bushels  more  cost 5,75 

3  bushels  cost 2,87-^ 

15  bushels  cost $14,37-^- c^;-25. 

In  New  York  currency,  8  yards,  &e.,  cost  as  many 
dollars  and  parts  of  a  dollar,  as  one  yard,  pound,  &;c. 
costs  shillings  and  parts  of  a  shilling. 


N 


146  ARITHrilETIC. 


Example  1.  At  4  shillings  3  pence  per  gallon,  N.  Y. 
currency,  what  will  12  gallons  cost? 

8  gallons  will  cost .......:.  $4,25 

4  gallons  will  cost 2,12^ 


12  gallons  will  cost $6,37^  Ans. 

Example  2.  At  3  shillings  6  pence  per  bushel,  N.  Y. 
currency,  what  will  31  i  bushels  cost? 

8  bushels  will  cost $3,50 


4X8=32  bushels  will  cost $14,00 

\  a  bushel  is  yV  of  8  bushels  ;  hence  yV  of  3,50,  which 

s  10  '  lb'' 

is  22  cents  nearly,  must  be  deducted.       Ans.  $13,78. 

[Note.   To   notice  very  small  fractions  of  a  cent  in 
such  problems  as  these,  is  being  more  nice  than  wise.'] 

1.  What  sum   must  be   paid  for  19  firkins  of  butter, 
each  containing  42  pounds,  at  16|  cents  per  pound  ? 

Ans.  $133. 

2.  At  12^2  cents  per  pound,  what  must  be  paid  for  4 
boxes  of  sugar,  each  containing  125  pounds  ? 

Ans.  $62,50. 

3.  What  must  be  paid  for  a  cask  of  rice,  containing  2 
cwt.  1  qr.  17  lb.  at  5  cents  per  pound?    Ans,  $13,45. 

Run  out  the  items  of  the  following  bills  :  the  first  is  N. 
England,  the  second  N.  York  currency. 

Mr.  Charles  Lardner, 

Bought  of  James  Fisher, 
1840.  5.    d. 

Jan.  3,     9  yds.  Bleached  cotton,  at  •  •    1-3 

6  lbs.  Souchong  tea  •  •  "  •  •  4  4 
12  «  Black  do.  .  . "  .  .  3  9 
18    "      Coffee "   •  •    1     3 

7  yds.  Irish  linen  ..."..  4  8 
24    "      Calico "   .  .    1     8 


North  Hampton,  Jan.  3,  1840. 
Received  payment, 


$29,57 
James  Fisher. 


PRACTICE. 


147 


Mr.  John  Schoolcraft, 


To  C.  P.  Smith,         Di 


1844. 

s. 

d. 

Jan.  8. 

To   10  lbs. 

Coffee, ...  at  . 

I 

9 

a 

"     12    " 

Suirar.  .  .   .at   . 

.    1 

3 

"    12. 

"     18    " 

Sabnoii  ...  at   • 

6 

Feb.  1. 

»      9    " 

Black  tea  .  •  at   • 

5 

3 

(4 

"    24    " 

Buckwheat .  at  . 

3 

*«      6. 

u      9    <. 

Corn'd  beef  at  • 

4 

New  York,  March  4,  1844. 

Received  payment, 


$12,21 


C.  P.  Smith. 


Lexington,  June  3,  1840. 
Mr.  William  Davis, 

Bought  of  Samuel  Merchant, 
*  $    cts. 

25 'j  lbs  Imperial  tea,  at 1 


1 5  pair  worsted  hose,  at 0 

7*  dozen  women's  gloves,  at.  .  -5 

16  dozen  knives  and  forks,  at  ...  1 
2',  cwt.  brown  sugar,  at 7 

'25  lbs.  Loaf  sugar,  at 0 


75 

87^ 

.37^ 

75" 

50 

30 


Received  payment  in  full. 


$152,31 


Samuel  Merchant. 


The  following  problems  are  a  compound,  or  rather 
combination  of  reduction  and  practice,  yet  all  strictly 
practical.  They,  moreover,  serve  to  explain  the  modern 
method  of  canceling,  which  is  no  less  useful  than  beauti- 
ful. 

1.   What  is  the  value  of  a  piece  of  gold  'weighing  1  lb. 
3  pemiyweights,  at  12^  cents  per  grain?     „in6-.  $729. 
243  dwt. 
M  3 


148 


ARITHMETIC. 


2.  At  5  cents  a  pound,  what  will  6  hundredweight  1 
quarter  cost  ?  Ans.  $35. 

5 

3.  At  $2,25  a  quarter,  what  will  1  ton  2  hundredweight 
cost?  ^^ns.  $198. 

22  cwt. 


4 


9 


cost? 

7  lbs. 

8  oz. 

=^n 

=  V  lbs. 

i 

^0  ^00 

4.  At  5  cents  per  ounce,  what  will  7  pounds  8  ounces 

Therefore, 

1^   3 

/l^  2 

But,  says  an  objector  to  this, mode  of  operation,  it  is 
very  fortunate  that  this  happened  to  be  8  ounces,  just 
half  a  pound  ;  what  would  you  have  done  had  it  been  9 
or  7,  or  10  oz.  ?  Compute  it  just  as  we  have 
at  8  ounces  ;  and  if  it  v/ere  9  ounces,  it  would  be  5  cents 
more,  as  it  is  5  cents  per  ounce;  if  2  ounces  more,  10 
cents  ;  if  1  ounce  less,  5  cents  less,  &c.:  no  difficulty 
with  a  willing  disposition. 

5.  What  will  16  yards  2  quarters  of  cloth  cost,  at  12^ 
cents  a  nail?  '  .'t??i5.  $33. 

66  qrs. 


V 


8 


.  How  many  coat  patterns,  each  containing  3  yards 
2  quarters,  can  he  cut  out  of  a  piece  of  cloth  containhig 
140  ells  Flemish?  ^^ns.  30. 

14      ^^^ 
^^         3 

7.   What  will  1  hogshead  of  wine  cost,  at  6^-  cents  a 
gill?  Ans.  $1^6. 

Observe,  that  6'^  •■=-/?  ^"^'  as  4  is  a  divisor  to  25,  it 
must  be  put  on  the  opposite  side  of  the  line. 


PRACTICE- 


149 


100 
4 


63 
4 
2 
4 

25 


or, 


16 


63 
4 
2 
4 


8.  Suppose  a  hogshead  of  inolasses,  wliich  cost  2'i 
dollars,  be  retailed  at  12^  cents  a  quart,  wliat  is  the  pro- 
fit on  it  ? 

63  gallons.  Sale,  31,50 

^  ^  Cost,  23 


2  ^ 


Profit,     8,50 
9.   What  will  5  barrels  of  flour  cost,  at  3.',   cents  per 
pound?  .6lni.  ^'Si,SO. 

10.  How  many  times  is  |  of  a  pint  contained  in  |  of  a 
gallon  ? 


or. 


.'his.  6|. 


20 

6i; 


We  have  already  remarked,  that  denominators  of  frac- 
tions must  go  over  the  line  from,  the  term  to  ivhich  they 
belong. 

1 1.  How  many  times  can  a  vessel,  holding  y^  of  a  quart, 
be  filled  from  ^  of  a  barrel,  containing  3U  gallons^ 


3U 


^/i  7 
10 


Ans.  46|. 


12.  If  1  acre  and  20  rods  of  ground  produce  45  bushels 
of  wheat,  at  that  rate,  how  much  will  9  acres  produce  ? 

Jlns.  360. 

la.  20r.  =  180.    ^0/1^0     /l^C)  8       45X8=360. 
45 

N.  B.  We  shall  plan  no  more  problems.  The  fol- 
lowing require  no  real  labor.,  save  correct  reasoning. 
When  the  numbers  are  properly  arranged,  a  few  clips 
with  the  pencil,*  and  perhaps  a  trijiing  multiplication, 

will  suffice. 

— 


150  ARITHMETIC. 


13.  At  H  cents  a  gill,  how  many  gallons  of  cider  can 
be  bought  for  $24  ?  Ans.    50. 

14.  How  many  casks,  each  containing  12  gallons,  can 
be  filled  out  of  half  a  tun  of  wine?  Ans.  \^\. 

15.  A  man  retailed  6  barrels  of  ale,  and  received  for  it 
$129,60 ;  at  what  price  did  he  sell  it  a  pint? 

An8. 1\  cents. 

16.  How  much  butter,  at  9  cents  per  pound,  will  pay 
for  12  yards  of  cloth,  at  2  dollars  19  cents  per  yard? 

.8ns.  2921b. 

17.  At  45^  dollars  per  acre,  what  will  32  rods  of  land 
come  to?     "  Am.  $9,10. 

18.  How  long  must  a  laborer  work,  at  62  ^^  cents  a  day, 
to  earn  $25  ?  Jins.  40  days. 

19.  A  merchant  sold  275  pounds  of  iron,  at  5|  cents  a 
pound,  and  took  his  pay  in  oats  at  50  sents  a  bushel ; 
how  many  bushels  did  he  receive  ?  Jins.  28|. 

20.  How  many  yards  of  cloth,  at  $4,66  a  yard,  must 
be  given  for  18  barrels  of  flour,  at  $9,32  a  barrel? 

Ans.  36. 

21.  How  long  will  it  require  one  of  the  heavenly  bo- 
dies to  move  through  a  quadrant,  at  the  rate  of  43'  12" 
per  minute  ?  Ans,  2^^^  hours. 

22.  If  a  comet  move  through  an  arc  of  7°  12'  per  day, 
how  long  would  it  be  in  passing  an  arc  of  90°  ? 

Ans.  VZ\  days. 

23.  What  is  the  cost  of  8  hogsheads  of  wine,  at  5  cts. 
per  pint?  .^/?5.  $201,60. 

24.  There  are  30  J^  square  yards  in  one  perch  of  land ; 
how  many  perches  are  there  in  363  square  yards  ? 

Am.  12. 

25.  What  will  18|  yards  cost,  at  75  cents  per  yard? 

Am.  $14,06' 

26.  If  10  persons  receive  1032  dollars  for  43  days' 
work,  how  much  does  each  man  earn  per  day? 

Ans.  $1,50. 

27.  How  many  times  will  a  wheel,  which  is  12  feet  in 
circumference,  turn  round  in  going  2  miles  ? 

Am.  880. 
28    What  number  of  revolutions  will  a  measuring  wheel 


SIMPLE     INTEREST.  151 


of  18  feet  6  inches  in  circiimferenoe,  make,  in  running 
150  miles?  Mns.  42810,  and  5  yd.  left. 

29.  How  many  times  will  a  wheel,  which  is  9  feet  2 
inches  in  circumference,  turn  round  in  going  65  miles? 

^ns.  37440. 

30.  If  a  man  can  travel  39  miles  20  rods  a  day,  how 
long  would  it  take  him  to  walk  round  the  globe,  the  dis- 
tance being  about  25000  miles  ?  ^ns.  640  days. 

31.  How  many  steps,  of  30  inches  each,  must  a  person 
take  in  walking  21  miles?  ^0?is.  44352. 

32.  How  many  suits  of  clothes,  each  containing  5  yards 
1  quarter,  can  be  cut  out  of  168  yards  ?  Jins.  32. 

33.  How  many  coat  patterns,  each  containing  2  yards 
1  quarter  1  nail,  can  be  cut  out  of  39  yards  1  quarter  1 
nail?  .^ns.  17. 

34.  How  many  times  can  a  vessel,  holding  4  gallons  2 
quarts,  be  filled  out  of  a  hogshead  of  wine  ?  ^7is.  14. 

35.  A  man  distributed  £3  3  s.  9  d.  among  a  company 
of  boys,  giving  each  3  s.  9  d.;  how  many  boys  were 
there  in  company?  ^^us.  17. 

N.  B.    The  last  four  problems  correspond  to   Note 
under  Art.  22. 


SIMPLE  INTEREST. 

(Art.  89.)  Simple  Interest  is  a  sum  allowed  for  the 
use  of  money,  goods,  or  property. 

The  sum,  for  which  interest  is  allowed,  is  called  the 
principal. 

The  rate  of  allowance  for  100  dollars,  for  1  year,  is 
called  the  rate  per  cent. — per  cent  meaning  per  100. 

The  amount  is  the  principal  and  interest,  added  to- 
gether.* 

*  The  rate  of  interest  is,  in  almost  all  countries,  established  by 
law ;  and  asking,  or  taking  more  than  such  rates,  is  termed  usnry, 
and  is  subject  to  legal  penalties.  In  New  England  and  Ohio,  the 
legal  rate  is  6  per  cent,,  in  New  York  7;  but. in  many  countries  of 
Europe,  money  can  be  obtained  at  4  and  5  per  cent. 


152  ARITH3IETIC. 


Every  question  of  simple  interest,  is  propcrlv  an  ex- 
ample in  proportion,  or  the  Rule  of  Three.  Three  terms 
are  oiven  to  tind  a  fourth  term.  'J'hus:  let  it  be  required 
to  tind  the  interest  of  any  sum,  say  $240.  for  1  year,  at 
6  per  cent.  The  term  6  per  cent,  means  that  100  dollars 
yields  6  dollars  in  one  year;  and  240  dollars  must  yield 
in  that  proportion. 

Hence,  as  100  :  240  : :  6  :  to  a  fourth  term, 
6 

14,40 
This  gives  $14,40  for  one  year's  interest;  and  if  tlie 
interest  were  required  for  two  or  three,  or  more  years, 
we  should  multipl}^  this  sum  by  2,  3,  or  more,  as  the 
case  may  require;  hence  v/e  have  the  following 

General  Rule.  Multipli^  the  principal  by  the  rate 
per  ce7if.,  and  that  product  by  the  time,  (a  year  being 
the  unit,)  and  divide  by  100. 

Let  the  pupil  remember  that  this  is  a  general  and  uni- 
versal rule,  equally  applicable  to  any  per  cent,  or  any  re- 
quired time,  and  all  other  rules  must  be  reconcilable  to  it; 
and,  in  fact,  all  other  rules  are  but  modifications  of  this. 

(Art.  90.)  When  the  interest  is  Q per  cent.,  we  7nay 
midliply  the  principal  by  half  the  ninnber  of  months, 
and  divide  by  100.  Let  us  reconcile  this  rule  with  the 
former.  What  is  the  interest  of  $800  for  16  montlis,  at 
6  per  cent.?  As  a  year  is  the  irnit  of  time,  we  must  di- 
vide the  16  by  12,  to  bring  it  into  years  and  parts  of  a 
year;  or  the  result  must  be  the  same,  if  we  multiply  by 
16  and  divide  by  12  (Art.  24).  Hence,  by  the  General 
Rule,  and  under  the  form  of  Art.  24,  we  have 

300= principal. 


12 

100 


6=rate. 
16= time. 


Now  cancel  6  in  12,  and  the  quotient  2,  into  16;  it  wil 
then  stand  thus: 

300 


100 


8  Ans.  $24. 


SIMPLE     INTEREST.  153 


That  is,  midliply  the  princijml  by  half  the  number  of 
months,  and  divide  by  100. 

(Art.  91.)  When  the  rate  of  interest  is  4  per  cent. 
midliply  the  principal  by  ^  the  number  of  months,  and 
divide  by  100. 

Let  us  investigate  this  rule.  What  is  the  interest  of 
$720,  for  2  years  and  G  months,  at  4  per  cent.?  Take 
the  general  rule :  thus, 

720 


100 


.4 
^^  months. 


10  =  1  the  months. 


When  thus  far  canceled,  the  remaining  part  of  the  op- 
eration, put  in  words,  is  the  rule  above  cited.  In  iJiis 
way,  we  can  show  that  3  per  cent,  will  require  us  to 
multiply  by  \  of  the  number  of  months,  and  divide  by 
100  ;  2  per  cent,  corresponds  with  ^  of  the  number  of 
months;  and  1  per  cent,  with  y'^  of  the  number  of 
months,  &lq. 

(Art.  92.)  When  the  per  cent,  is  complex  or  mixed, 
as,  4|,,  5|,  7|,  &c.,  it  may  be  most  convenient  to  first 
cast  the  interest  at  1  per  cent.,  by  multiplying  the  prin- 
cipal by  -j'j  of  the  number  of  months,  and  dividing  by 
100 ;  and  afterwards,  multiplying  by  the  per  cent.,  what- 
ever it  may  be  ;  or  we  may  proceed  by  the  general  rule. 
The  first  will  be  most  brief,  if  the  months  will  divide  by 
12,  or  if  we  find  any  large  common  factors,  between  100 
and  the  months. 

N.  B.  When  days  are  in  question,  they  must  be  con- 
sidered as  parts  of  a  month,  30  days  to  a  month  ;  there- 
fore, 3  days  is  ,1  of  a  month,  6  days  ,2,  9  days  ,3,  &;c. ; 
or,  in  vulgar  fractions,  1  day  is  -3'^,  2  days  y^,  &c. 

EXAMPLES. 

1.  What  is   the  interest  of  $75,  for  1  year  6  months 
and  12  days,  at  6  per  cent.  ?    • 
y.    m.    d. 
1     6     12  =  18,4  months,  half  9,2. 


154  ARITHMETIC. 


By  rule  (Art.  00)— 

■ji   3  4)27,6 


4 


9,2 


$6,90  ^ns. 

2.  Find  the  interest  of  $467,17  for  3  years  5  months, 
at  7  per  cent. 

This  had  better  be  done  by  the  general  rule. 

467,17 

7 


Interest  for  1  year, 32,7019 

3 


Interest  for  3  years, 98,1057 

Interest  for  j  of  a  year,  or  4  months,    .  .  .  10,9006 
Interest  for  1  month,  or  j^  of  4  months,    •  •    2,725 


.^ns. $111,731 

3.  Required  the  interest  of  $64,50  for  3  years  5  mon. 
and  10  days,  at  7  per  cent.? 

The  days  here  make  the  operation  more  brief,  if  we 
take  the  canceling  form,  and  cast  the  interest  at  1  per 
cent.  (Art.  92.) 


Months,  41i  =  ^r.  3  ^^ 

100 


^^,li6  21,50 


21,50 
31 

21,50 
645 

300)666,50 


Interest  at  1  per  cent.  ....   2,2225 

7 


Interest  at  7  per  cent.    .  .  .  15,5575  or  $15,56  r^ns. 


SIMPLE    INTEREST.  155 


4.  What  is  the  interest  on  $125,  for  17  days,  at  6  per 
cent.? 

First  find  the  interest  for  15  days,  i-  of  one  month, 
(Art.  90.) 


4 

m 

4 


^^A  i5 


That  is  y^g  of  a  dollar,  or  ,31'^  cents  is  the  interest  for  15 
days  ;  /^  of  this  in  addition  will  give  the  interest  for  17 
days.  Ans.  35  cents  4  mills. 

5.  AVhat  is  the  interest  of  $204,  for  40  days,  at  6  per 
cent.? 

40  days  =  l|-  month=|^,  half  =f . 

^     m  68 

100       "^ 

Jins  $1,36. 

6.  Find  the  interest  on  $420,  for  2  years  1  month  and 
27  days,  at  6  per  cent. 


Time  25,9  months,     p, 
100 


^^0  21 
25,9 
25,9 
21 


259 
518 


10)5439 

$54,39  Ans. 

7.  What  is  the  amount  of  $560,  for  9  months  and  20 
days,  at  6  per  cent.?  .dns.  $587,066+ 

8.  What  is  the  interest  of  $50,11,  for  1  year  and  21 
days,  at  6  per  cent.?  Ans.  $3,18  +  . 

9.  What  is  the  interest  of  $340,50,  for  3  months,  at  6 
percent.?  e^n.s.  $5,1 1. 

10.  What  is  the  interest  of  $180,  for  1  year  2  months 
and  6  days,  at  6  per  cent.?  Ans.  $12,78. 

11.  What  is  the  interest  of  25  dollars,  for  3  years  6 
months  and  20  days,  at  6  per  cent.?         A7is.  $5,33^. 


156  ARITHMETIC. 


12.  What  is  the  interest  of  125  dollars,  for  5  years  5 
months  5  days,  at  6  per  cent.?  '       M)i.s.  40,73. 

13.  Wliat  is  the  amount  of  56  dollars,  for  9  years  8 
months  3  days,  at  6  per  cent.?  Jins.  $88,51. 

14.  What  is  the  amount  of  84  dollars,  for  5  years  5 
months  and  9  days,  at  5  per  cent.?  ^^ns.  $106,85. 

15.  What  is  the  interest  of  $72,05,  for  3  montlis,  at  2 
per  cent.?  e/??i5.  36  cents. 

16.  What  will  625  dollars  produce  in  3  years  2  months 
and  7  days,  at  10  per  cent.?  .^?75.'$199,13-[-. 

17.  How  much  interest  will  $865,25  produce  in  9 
years  4  months  and  15  days,  at  6  per  cent.? 

Jns.  $486,70+. 

18.  What  is  the  interest  on  9000  dollars,  for  1  year  4 
months  and  8  days,  at  5  per  cent.?  ^ns.  $610. 

19.  What  is  the  interest  on  998  dollars,  for  3  years  10 
months  and  18  days,  at  6  per  cent.?  Compute  for  4 
years,  and  deduct  the  interest  of  the  sum  for  1  m.  and  12  d. 

Thus, ...  998         For  the  days, 998 

,24         ^  the  number  of  months,     ,7 


1996  Deduct $2,986 

3992 


$239,52 


(Art.  93.)  When  the  principal  is  in  pounds,  shillings 
and  pence,  the  pound  being  the  unit,  reduce  the  shillings 
and  pence  to  the  decimal  of  a  pound,  by  Art.  69. 

20.  What  is  the  interest  of  £75  12  s.  6d.  for  2  years 
4  months  and  10  days,  at  5  per  cent.? 


12 
20 


6  , 
12,5 


Form, ^         12 

3 


Principal=75,625 
Time  28J-  months. 

^^,0^     3,25 


(4  100 

1  '1 


5 

85 


SIMPLE    INTEREST. 


157 


3,25 
5 

16,25 

85 

8125 
13000 

£ 

144)1381,25(9,599  nearly. 
1296      20 


852   11,980 
720     12 


1325  11,76 
1296 


1290 


£     s.       d. 
Ans.   9  11  IH. 


2 1 .   Find  the  amount  of  £320  8  s.  for  ten  years,  at  6 
per  cent. 


O  A.  —  2  0 — To  — »^ 

£  S.  cl. 
Principal,  320  8  0 
Interest,  .192     4     91 


100 


Amount,  .  512   12     9i 


320,4 
60 
320,4 
6 

£192,24 
20 

4,80 
12 


Form,     100 


9,60 

22.  Find  the  amount  of  £1230  4  s.  for  8  years  and 
6  months  at  4  per  cent.  ? 

1230,2 

34  .^ns.  £1648  9  s.  4  d. 

34  being  \  of  the  number  of  months,  corresponding  to  4 
per  cent.  (Art.  91). 

23.  What  is  the  interest  of  £21  18  s.  4  d.  for  3  years 
and  4  months,  at  7  per  cent.?  Ans.  £5  2  s.  6^'d. 

O 


158  ARITH3IETIC. 


24.  What  is  the  amount  of  £156  9  shillings  3  pence, 
for  1  veai*  and  9  months,  at  6  per  cent.? 

Ans.  £172  17  s.  9d.  3  qr.  + 

25.  What  is  the  amount  of  £27  2  shillings  6  pence  3 
quarters,  for  1  year  and  10  months,  at  6  per  cent.? 

Am.  £30  2s.  2d.  3  qr.-|- 

26.  What  is  the  interest  of  £75  10  shillings,  for  2 
years  and  3  months,  at  6  per  cent.? 

.^715.  £10  3s.  lOd.-l- 

27.  What  is  the  interest  of  450  dollars,  for  6  months 
and  20  days,  at  ^\  per  cent.? 

Per  cent.   5^=^.     Time,  6|  months=\?,  years=§^ 


2 
By  the  General  Rule,  2  ^d 

Art.  89, .  .• A00 


m  50 
11 

\6  4)55 


Ans.  13,75 


(Art.  94.)  When  interest  is  required  on  any  sum  for 
days  only,  it  is  a  universal  custom  to  consider  30  days  a 
month,  and  12  months  a  year;  and,  as  the  unit  of  time 
is  a  year,  the  interest  of  any  sum  for  one  day  is  ^^^, 
what  it  would  be  for  a  year.  For  2  days,  ^^f^,  &c.;  hence, 
if  we  multiply  by  the  days,  we  must  divide  by  360. 


EXAMPLES. 

1.  What  is  the  interest  of  $327,30,  for  25  days,  at  6 
per  cent.? 

100 


327,30")  Gives  interest  for  1  year,  by  General 
6       5  Rule,  Art.  89. 
25 

The  6 'will  cancel  in  the  360:  then  the  operation  will 
stand  thus : 


360 


100 
00 


327,30 
25 


This  gives  the  following  special  rule,  when  the  inter- 
est is  six  per  cent.,  and  the  time,  days. 

Rule.  Multiply  the  principal  by  the  number  of  days; 
divide  the  product  by  60,  and  remove  the  decimal  point 
two  places  to  the  left. 


SIMPLE    INTEREST.  159 


To  obtain  the  result  in  this  particular  example,  we  ob- 
serve the  numbers  will  still  further  cancel. 

2.  What  is  the  interest  of  854  dollars,  for. 30  days,  at 
6  per  cent,  per  ainuiin  ?  Ana.  $4,27. 

3.  AVhat  is  the  interest  of  1100  dollars,  for  48  days,  at 
6  per  cent,  per  annum  ?  Jlns,  $8,80. 

4.  What  is  the  interest  of  3459  dollars,  for  75  days, 
at  6  per  cent,  per  annum  ?      .^n.?.  $43,23  cts.  7  m.-4- 

5.  What  is  the  interest  of  1500  dollars,  for  60  days, 
at  5  per  cent,  per  annum?  Ans.  $12,50. 

6.  WHiat  is  the  interest  of  1000  dollars,  for  29  days  ? 

Ans.  $4,833  -f- 

7.  What  is  the  interest  of  204  dollars,  for  40  days  ? 

Ans.  $1,36. 

8.  W^hat  is  the  interest  of  472  dollars,  for  18  days  ? 

Ans.  $1,416. 
When  no  rate  is  mentioned,  6  per  cent,  is  understood. 

(Art.  95.)  To  compute  interest  on  Notes,  Bonds, 
and  Mortgages,  on  which  partial  payments  have  been 
made,  two  or  three  rules  are  given.  The  first  of  the  follow- 
ing is  called  the  common  rule,  and  applies  to  cases  where 
the  time  is  short,  and  payments  made  within  a  year  of 
each  other.  This  rule  is  sanctioned  by  custom  and  com- 
mon law  ;  it  is  true  to  the  principles  of  simple  interest, 
and  requires  no  special  enactment.  The  other  rules  are 
rules  of  law,  made  to  suit  such  cases  as  require  (either 
expressed  or  implied)  annual  interest  to  be  paid,  and  of 
course  apply  to  no  business  transactions  closed  within  a 
year. 

Rule  1.  Compute  the  interest  of  the  principal  sum 
for  the  tvhole  time  to  the  day  of  settlement,  and  find  the 
amount.  Compute  the  interest  on  the  several  payments 
from  the  time  each  was  paid  to  the  day  of  settlement ; 
add  the  several  payments  and  the  interest  on  each  to- 
gether, and  ccdl  the  sum  the  amount  of  the  payments. 
Subtract  the  amount  of  the  payments  from  the  amount 
of  the  principal,  ivill  leave  the  surn  due. 

The  second  rule  which  follows,  is  the  rule  of  the  Su- 
preme Court  of  the  United  States,  and  with  some  trifling 


1 60  arith:\ietic. 


variation,  adopted  by  several  of  the  States.  We  give  it 
in  the  language  of  Chancellor  Kent,  as  expressed  in  the 
New  York  Chancery  Reports  : 

Rule  2.  '' .Hpply  the  payment,  in  the  first  place,  to 
the  discharge  of  the  interest  then  due.  If  the  paipntnt 
exceeds  the  interest,  the  surplus  goes  toivnrds  discharg- 
ing the  principal,  and  the  subsequent  interest  is  to  be 
computed  on  the  balance  of  principal  remaining  due. 
If  the  payment  be  less  than  the  interest,  the  surplus  of 
interest  must  not  be  take?!  to  augment  the  principal ; 
but  interest  continues  on  the  former  principal,  until  the 
period  when  the  payments  taken  together  exceed  the  in- 
terest due,  and  then  the  surplus  is  to  be  applied  toward 
discharging  the  principal ;  and  interest  is  to  be  compu- 
ted on  the  balance,  as  aforesaid.''^ 

Rule  3d,  which  follows,  is  one  adopted  by  the  Su- 
preme Court  of  the  State  of  Connecticut,  is  more  expli- 
cit and  guarded  against  granting  partial  compound  inter- 
est, than  the  Chancellor's  rule.  The  Connecticut  rule 
does  not  require  interest  to  be  paid  until  the  principal  has 
run  a  year.  It  supposes  that  interest  is  not  due  under 
that  time,  though  payments  may  have  been  made  every 
few  months,  and  the  debtor  is  allowed  interest  on  his  pay- 
ments up  to  the  end  of  the  year,  agreeably  to  our  first 
rule.     But  let  the  rule  speak  for  itself: 

CONNECTICUT  RULE. 

Rule  3.  ''Find  the  amount  of  the  principal  to  the 
time  of  the  first  payment,  if  it  be  a  year  or  more  after 
interest  commenced,  and  subtract  this  payment  from  it. 
The  remainder  will  be  a  neiv  principal,  on  which  cal- 
culate the  amount  to  the  jiext  payment,  and  from  it  sub- 
tract the  payment.  Proceed  in  this  way  from  payment 
to  payment,  until  they  are  all  employ ed, provided  a  year 
or  rnore  intervenes  between  each  two  payments.  But, 
if  the  intervening  time  be  less  than  a  year,  find  the 
amount  of  the  last  principal  for  a  year,  and  of  the  pay- 
ment up  to  the  same  time,  and  subtract  the  latter  fr on 
the  former.  If  however,  a  year  overruns  the  time  of 
settlement,  find  the  amounts  up  to  that  time,  instead  of 
for  a  year. 


SIMPLE     INTEREST.  161 


If  any  payment  is  less  than  the  interest  then  due^ 
continue  the  former  principal  to  the  next  payment,  and 
put  both  payments  together. 

EXAMPLES. 

1.  For  value  received,  I  promise  to  pay  Smith  Jones, 
or  order,  one  thousand  dollars  on  demand,  with  interest 
from  date.  William  Smith. 

Boston,  January  1st,  1836. 

On  taking  up  this  note,  Deo.  1st,  1836,  the  following 
endorsements  were  found  on  the  back: 
*1.      1836.  March  1.    Received, $250. 

2.  1836.  July      1.     Received, 300. 

3.  1836.  Oct.       1.    Received, 125. 

What  was  due  on  the  note?  ,/ins.  By  Rule  1,  $360. 

By  Rule  2,     360,90. 

It  will  be  observed,  however,  that  this  note  should  be 
settled  by  Rule  1.  To  show  the  harmony  and  discrep- 
ancy of  these  rules,  let  us  suppose  the  following  case : 

2.  A  gave  his  note  to  B,  for  10,000  dollars  ;  at  the  end 
of  4  months,  A  paid  6000  dollars;  and  at  the  expiration 
of  another  4  months,  he  paid  an  additional  sum  of  3000 
dollars  ;  how  much  did  he  owe  B,  at  the  close  of  the 
year? 

By  the  Common  Rule, 

Principal, $10,000 

Interest  for  the  whole  time, 600 

Amount, $10,600 

1st  payment,.  •  .  $6000 
Interest,  8  months,  240 
2d  payment,  3000 

Interest,  4  months,       60 


Amount, $9300  9,300 


Due, $1300 

It  will  be  observed  that  the  Connecticut  rule  must  work 


*  We  find  the  time  elapsed    between  two  payments,  by   com- 
pound subtraction,  (Art.  29.) 

o2 


162  ARITHMETIC. 


this  question  liie  same  way,  and  arrive  at  the  eame  re- 
sult.    Now  let  us  take  the"2d,  or  Chancellor  Kent's  rule. 

Principal,     $10,000 

Interest,  4  months, 200 

Amount, 10,200 

Subtract  payment, 6,000 

New   principal, 4,200 

Interest,  4  months, 84 

Amount, 4,284 

Second  payment, 3,000 

New  principal, 1,284 

Interest,  4  months, 25,68 

Remains  due,     $1,309,68 

The  dilTerence  resulting  from  the  different  rules,  as 
applied  to  this  case,  is  $9,68.  To  thoroughly  under- 
stand this,  let  us  consider  that  $200  interest  was  paid  to 
B.,  out  of  the  6000  dollars,  8  months  before  it  was  pro- 
perly due  ;  and  B.  had  the  opportunity  to  put  that  sum 
out  at  interest  again,  and  thereby  receive  partial  com- 
pound interest.  This  is  the  reason  that  the  Connecticut 
rule  requires  interest  to  run  a  year  on  the  principal,  and 
interest  on  the  payments  up  to  the  end  of  a  year.*  That 
this  is  a  correct  view  of  the  case,  let  us  cast  the  interest 
on  the  $200  for  8  months,  and  on  the  $84  for  4  months, 
which  was  paid  out  of  the  $3000,  4  months  before  it 
was  due.     Interest  on  200  dollars,  8  months,=8. 

"        on    84  dollars,  4  months, =1,68, 


$9,68. 


*NoTE.  Annual  mieresi,  actually  paid  when  due,  is  in  fact  [or 
in  effect]  compound  interest,  and  these  State  rules  seem  to  be  a 
sort  of  compromise  between  simple  and  compound  interest ;  they  re- 
cognize the  principles  of  simple  interest,  when  they  sutler  interest  to 
overrun  a  year,  and  give  compound  interest,  liy  not  allowing  interest 
on  the  payments,  or  by  requiring  interest  to  be  paid  until  the  day  of 
I  settlement. 


SIMPLE     INTEREST.  163 

3.  A  note  for  2000  dollars  was  given  January  4,  1827, 
on  which  there  were  the  lollowing  payments  : 

February  19,   1828, $400. 

June  29,  1829, 1000. 

November  14,  1829, .520. 

How  much  remained  due  Dee.  24,  1830,  interest  at  6 
per  cent.? 

Principal, $2000 

Interest  from  Jan.  4,  '27,  to  Feb.  19,  '28, 

(13^   months,) 135 

First  amount, 2135 

First  payment, 400 

Balance,  forming  a  new  principal, 1735 

Interest  from  Feb.  19,  '28,  to  June  29,  '29, 

(16^   months,) 141,69 

Second  amount, 1876,69 

Second  payment, 1000 

Balance,  forming  a  new  principal, 876,69 

Interest  from  Ju"ne  29,  to  Nov.  14,  (4|  mos.)       19,72 

Third  amount, 896,41 

Third  payment, 520 

Balance,  forming  a  new  principal, 376,41 

Interest  from  Nov.  14,  to  Dec.  24,  (131  m.)         25,09 

Balance  due  on  taking  up  the  note, $  401,50 

This  is  by  the  United  States  Court  Rule,  which  is 
sometimes  called  the  New  York  and  Massachusetts  rule 
(Rule  2).     By  the  Connecticut  rule,  the  balance  due  on 

this  note  is $401,16. 

The  difference,  as  the  reader  will  observe,  arises  by  pay- 
ing (according  to  the  New  York  rule)  $19,72  interest, 
71  months  before  the  Connecticut  rule  considers  it  due  ; 
and  by  the  Connecticut  rule,  we  cast  interest  on  $870,69 
for  one  year,  and  then  on  the  payment  of  $520,  for  7J- 


164  ARITIIMETIC>^^\yx\.^^ 


months,  up  to  the  end  of  the  year,  aiicl  subtract  the 
amount  for  the  amount  of  the  principal  sum.  Then  we 
cast  interest  on  the  remainder,  to  the  day  of  setllement. 
By  Rule  1,  no  interest  would  be  paid  until  the  day  of 
setdement;  hence,  with  heavy  sums  and  long  trjnsac- 
tions,  or  in  any  case  where  annual  interest  is  expected, 
Rule  1  is  not*  proper;  but  where  a  note  or  bond  does 
not  run  two  years,  and  several  payments  are  made  witli- 
in  a  few  months  of  each  other,  Rule  1  corresponds,  both 
in  letter  and  spirit,  to  the  principles  of  simple  interest, 
opposed  to  usury.  A  correct  understanding  of  this  mat- 
ter among  the  people,  may  prevent  considerable  litigation 
and  some  injustice/-' 

4.  $2500.'  One  year  from  date,  I  promise  to  pay  James 
Vance,  or  order,  two  thousand  tive  hundred  dollars,  with 
interest,  for  value  received.  George  Kidd. 

Bosloih  Jan'y  1,  1836. 

As  a  personal  favor  to  James  Vance,  George  Kidd 
paid  on  this  note, 

Mav  1,  1836, ^400. 

Julv  1,  1836, 300. 

No'vember  1,  1836, 600. 

February  1,  1837, 800. 

George  Kidd  not  being  able  to  pay  more  at  this  time, 
and  James  Vance  being  pressed  for  money,  disposed  of 
the  note  at  under  value,  to  a  note-shaver.  Kidd  took  up 
the  note  Julv  1,  1837;  what  was  then  due? 

Ans.  Bv  Rule  1, S535. 

By  Rule  2, 541,77. 

Difference, 6,77. 

A  difference  of  feeling  now  arose,  as  to  which  rule 
should  decide  the  balance  ;  the  note-shaver,  like  Shylock, 

*  On  making  inquiry  of  a  lawyer,  of  very  considerable  practice, 
how  he  made  up  balances  on  notes,  mortgages,  &c.,  cm  which  several 
payments  had  bten  made,  he  replied— When  my  cUent  is  plaintiff, 
I  go  from  payment  to  payment:  when  defendant,  and  is  to  pay  the 
balance,  I  reckon  on  the  whole  time,  and  then  on  the  payments  up 
to  the  day  of  settlement ;  and,  if  the  opposing  lawyer  makes  no  ob- 
jections, all  is  well ;  if  he  does,  I  must  conform  to  law. 


SIMPLE     INTEREST.  165 


contended  for  the  law,  the  time  being  more  than  a 
year.  'J'he  other  contended  that  he  paid  part  of  the  note 
before  it  was  due,  to  accommodate  tlie  former  holder, 
and  the  present  holder  bonirht  it  under  vahie ;  and  more- 
over, according  to  the  principles  of  equation  of  payments, 
no  partial  compound  interest  could  be  exacted.  Which 
rule  should  decide  the  balance?  The  technicalities  of  the 
law  favor  Rule  2d ;  the  true  spirit  of  the  law  would  es- 
timate by  Rule  1st. 

5.  For  value  received,  I  promise  to  pay  Joseph  Linn,  or 
order,  two  hundred  and  seventeen  dollars  and  tifty  cents, 
in  four  months,  with  interest  afterwards.       G.  Davis. 

Hartford,  July  11,  1831. 

On  tins  note  were  the  three  following  endorsements : 

1831.  Nov.  16, $93. 

18.32.  Feb.  12, 50. 

1832.  Aug.  2, 67,50. 

1832,  Oct.  4,  the  note  was  taken  up;  how  much  was 

due  upon  it?  ^n.5.  $11,05. 

This  being  in  Connecticut,  must  be  computed  by  the 
Connecticut  rule. 

6.  A  note  of  hand,  dated  New  York,  April  4,  1838, 
was  given  for  the  payment  of  six  hundred  dollars,  on 
which  there  were  endorsements  as  follows: — July  10, 

1838,  $84,64;  November  22,  1838,  $10,00;  April  30, 

1839,  $14;  December  5,  1839,  $309.  What  was  the 
balance  due  on  taking  up  the  note,  April  5,  1840  ? 

Ans.  $251,03. 

Neiv  York,  July  10,  1830. 

7.  Four  years  from  date,  we  jointly  and  severally  pro- 
mise, for  value  received,  to  pay  to  the  order  of  George 
R.  Guernsey,  eight  hundred  and  sixty-four  dollars,  with 
interest.  Robert  C.  Duncan. 

David  Johnston. 
On  this  note  are  endorsements  as  follows: 

April  6,  1831, $34. 

June  21,  1832, 300. 

February  26,  1833, 180. 

January  1,  1834, 40. 

What  was  the  balance  due  at  the  maturity  of  the  note  ? 


166  ARITHMETIC. 


COMPOUND  INTEREST. 

(Art.  96.)  When  interest  is  payable  at  definite  and 
stated  periods,  such  as  rents  and  special  loans,  for  a  deti- 
niie  time,  the  time  comes,  and  neither  principal  nor  in- 
terest are  paid — both  due.  A  new  obligation  throws 
them  into  one  sum,  forming  a  new  principal,  the  interest 
being  on  interest.  Another  period  elapses;  nothing  is 
paid,  and  a  like  obligation  takes  place,  interest  again  be- 
ing on  interest,  which  is  called  compound  interest.  This 
explanation  must  give  us  the  following 

Rule.  Compute  the  interest  up  to  the  time  that  it  be- 
came payable,  and  add  it  to  the  principal;  then  cast 
the  interest  on  that  amount  for  the  next  period,  and 
add  it  to  its  principal,  and  so  on.  The  first  principal 
subtracted  from  the  last  amount,  will  give  the  compound 
interest  for  the  whole  time. 

Example.  What  is  the  compound  interest  of  $250  for 
3  years,  at  7  per  cent.?  The  compound  interest  of  $250, 
is  250  times  greater  than  the  compound  interest  for  $1,00. 
We  therefore  compute  on  1  dollar. 

Principal  and  interest,  1st  year, 1,07 

Multiply  by  1+07,  or, 1,07 

7,49 
1,07 

Principal  and  interest,  2  years, 1,1449 

Multiply  by 1,07 

80143 
1,1449 


Amount  of  $1,  for  3  years, 1,225043 

Drop  $l,and  we  have  ,225043  for  the  compound  interest 
of  1  dollar;  which, multiplied  by  250,  gives  $56,36,  Ans. 
The  above  operation  might  be  expressed  thus: 
(1,07=^— 1)X  250  =  56,26. 


COMPOUND    INTEREST.  167 


2.  Find  the  amount  of  1  dollar  every  year,  for  4  years, 
at  compound  interest  at  6  per  cent. 

Amount  at  the  end  of  the  1st  year, $1,0G 

1,00 

Interest,  2d  year, ,00  3G 

Principal, 1,00 

Amount  at  the  end  of  2d  year, 1,12  30 

1,06 

Interest,  3d  year, ,0674  16 

Principal, 1,12  36 

Amount  at  the  end  of  4th  year, 1,19  10  16 

1,06 

Interest,  4th  year, ,07  14  60  96 

Principal, 1,10  10  16 

Amount  at  the  end  of  4th  year,  •  •  •  1,26  24  769 
(Art.  97.)  In  this  manner,  we  can  compute  the  amount 
of  one  dollar  at  any  per  cent.,  for  any  number  of  years, 
and  it  is  evident  that  two  dollars  will  amount  to  twice  as 
much  as  one  dollar;  three  dollars,  three  times  as  much  ; 
ten  dollars,  ten  times  as  much,  &c.  Hence,  if  we  have 
a  table  with  the  amount  of  one  dollar,  at  several  rates, 
and  for  a  number  of  years,  we  can  take  such  amount 
from  the  table,  and  multiply  it  by  the  number  of  dollars 
in  question,  and  the  product  will  be  the  amount  of  that 
number  of  dollars,  for  the  time  required. 

TABLE    1, 

Showing  flip  amovjif  of  One  Dollar  for  one  month  and 
quarters,  at  the  several  rates  mentioned. 


1  month. 

1  quarter. 

2  quarters 

3  quarters 

4  qr.  1  yr. 


4  per  ci'iii.4-i- per  cent.;  5  per  cent,  jo^  per  cent.  6percen 


1,003333 
1,009853 
1,019804 
1,029852 
1,04 


1,003750  1,0041G7  1,004583  1,005 
1,011065  1,012272  1,013475  1,014074 
1,022252  1,024694  1,027132  1,029563 
1,033563  1,037270  1,040973,1,044671 
1,045   11,05    1,055   |l,06 


168 


ARITHMETIC. 


TABLE    2,*' 

Showing  the  amount  of  One  Dollar,  from  one  to  twen- 
ty years. 


r  4  percent.  I  4^  per  cenl.   5  per  cent.  SJ  per  cent.   G  per 


1,0400000 
1,0816000 
1,1248640 
1,1698585 
1,2166529 
1,2653190 
1,3159317 
1,3685690 
1,4233118 
1,4802442 
1,5394540 
1,6010322 
1,6650735 
1,7316764 
1,8009435 
1,8729812 
1,9479005 
2,0258161 
2,1068491 
2,1911231 


1,0450000 
1,0920250 
1,1411661 
1,1925186 
1,2461819 
1,3022601 
1,3608618 
1,4221006 
1,4860951 
1,5529694 
1,6228530 
1,6958814 
1,7721961 
1,8519449 
1,9352824 
2,0223701 
2,1133768 
2,2084787 
2,3078603 
2,4117140 


1,0500000 
1,1025000 
1,1576250 
1,2155062 
1,2762815 
1 1,3400956 
1,4071004 
1,4774554 
1,5513282 
;  1,6238946 
1,7103393 
'1,7958563 
i  1,8856491 
:i,9799316 
2,0789281 
|2,1828745 
1 2,2920 183 
,2,4066192 
2,5269502 
2,6532977 


1,0550000 
1,1130250 
1,1742413 
1,2388246 
1,3069598 
1,3788426 
1,4546789 
1,5346862 
1,6190939 
1,7081440 
1,8020919 
1,9012069 
2,0057732 
2,1160907 
2,2324756 
2,3552617 
2,4848011 
2,6214652 
2,7656458 
2,9177563 


1,0600000 
1,1123600 
1,1910160 
1,2624769 
1 1,3382256 
1,4185191 
1,5036302 
11,5938480 
1,6894789 
11,7908476 
1 1,8982985 
2,0121964 
2,1329282 
1 2,2609039 
'2,3965581 
2,5403517 
12,6927727 
'2,8543391 
3,0255995 
|3,2071355 


*  By  inspection  of  Table  2,  it  will  be  seen  that  any  sum  doubles 
itself  at  4  per  cent.,  compound  interest,  in  less  than  18  years;  at  4^ 
per  cent.,  in  less  than  16;  at  5  per  cent.,  in  less  than  15  ;  and  at  6 
per  cent.,  in  less  than  12  years.  The  amount,  at  every  rate,  in- 
creases faster  and  faster,  as  the  time  increases.  In  19  years,  at  6  per 
cent.,  the  amount  is  more  than  3  times  the  original  amount ;  and  in 
40  years,  more  than  ten  times  that  sum.  These  facts  show  that  it 
would  work  much  injustice  and  ruin,  to  allow  compound  interest. 
Though  in  times  of  prosperity,  and  in  schemes  of  speculation,  mo- 
ney may  be  worth  6,  and  even  a  greater,  per  cent.;  yet,  in  the  sober 
realities  of  life,  in  unsettled  estates  in  law,  men  in  a  state  of  bank- 
ruptcy, or  other  contingency,  which  must  leave  debts  in  an  unsettled 
condition,  ruin  must  follow,  if  compound  interest  were  allowed  by 
law.  Capital  would  swallow  up  competency,  the  widow  and  the 
orphan  would  lose  their  inheritance,  and  the  unfortunate  could 
never  recover  from  misfortune.  There  is  much  small  logic  abroad, 
concerning  the  abstract  justice  of  compound  interest ;  but  this  sub- 
ject has  engaged  the  attention  of  the  most  gifted  minds,  in  all  ages, 
and  in  every  civilized  country ;  and  the  result  has  invariably  been 
a  condemnation  of  eorapound  interest 


COMPOUND     INTEREST.  109 


3.  Find  the  compound  interest  for  475  dollars,  for  5 
years,  at  5  per  cent.     From  the  table,  opposite  5  years, 

and  under  5  per  cent.,  we  find 1,34009 

For  the  amount  of  $1.     We  multiply  by 475 

6,70045 
9,38003 
5,30036 

Amount,  .  .  .  036,54275 
Principal,   •  .  475 


^ns.  161,54  + 
N.  B.  When  the  interest  alone  is  required,  as  in  the 
preceding  example,  neglect  one  unit  in  the  table,  and  ihe 
product  will  be  the  interest  at  once. 

4.  Wliat  will  786  dollars  amount  to,  in  3  years  and  a 
quarter,  at  6  per  cent.?  ^^ns.  $949,87  -{- 

l,1910ft)=amount  of  1  dol.  for  3  years. 
l,014674=am't  of  1  dol.  for  3  mos.  or  1  qr. 

4764064 
8337112 
7146096 
4764064 
1191016 
1191016 


1 ,208492968784=the  amount  of  $1  for  the  given  time. 
78.^=: the  given  sum. 


7250957812704 
9607943750272 
8459450781488 


949,875-}- =the  amount  for  the  time  required. 

5.  What  is  the  compound  interest  of  629  dollars,  for 
7  years,  at  6  per  cent.?  ^^ns.  $316,78  -f 

6.  What  will  be  the  compound  interest  on  a  bond,  for 
I  $750,  for  4|  years,  at  5^  per  cent.?    Ans.  $217,18  + 

— 


170  ARITHMETIC. 


7.  What  is  the  compound  interest  of  3200  dollars,  for 
8  years,  at  5}  per  cent.? 

8.  What  is  the  compound  interest  of  720  dollars,  for 
7  years,  at  6  per  cent.?  Ans.  $362,61  -\- 

9.  A  broker  loans  money  at  the  rate  of  6  per  cent,  per 
annum,  but  renews  quarterly ;  how  much  interest  does 
he  receive  per  annum,  on  100  dollars  ? 

Principal, 100 

Interest,  1st  quarter, 1,50 


Principal,  2d  quarter, 101,50 

Rate  per  quarter, ,015 


50750 
1015 

Interest,  2d  quarter, 1,52250 

Principal, 101,50 


Principal,  3d  quarter, 103,0225 

,015 


5151125 
1930225 


Interest,  3d  quarter, 1,5453375 

103,0225 


Principal,  3d  quarter, 104,56783 

,015 


52283915 
10456783 


Interest,  4th  quarter, 1,56851745 

Principal, 104,56783 


106,13634 
100 


Ans.  $6,136  -f 


DISCOUNT    AND    BANKING.  171 


On  1000  dollars,  the  interest  would  be  61  dollars  and 
30  cents.  It  will  be  observed,  that  this  problem  does  not 
correspond,  in  number,  to  Table  2  ;  nor  should  it,  though 
the  rate  per  annum  is  tbe  same;  the  imit  of  time,  in  that 
table  is  one  year,  and  the  unit  of  time  in  this  example,  is 
one  quarter. 


DISCOUNT  AND  BANKING. 

(Art.  08.)  Discount, — counting  back, — a  deduction 
from  an  account,  applicable  to  demands  due  at  a  future 
time  not  drawing  interest,  and  applicable  to  notes  on 
which  the  interest  is  paid  in  advance  ;  the  drawer  re- 
ceiving a  sum  in  hand,  which,  at  the  specified  rate  of  in- 
terest, will  amount  to  the  face  of  the  note  in  the  speci- 
(lecl  time.  The  problem,  then,  of  discount,  is, from  the 
raff,  lime  and  amount,  to  find  the  principal.  Interest 
is  proportion  (Art.  89) ;  hence  discount  is  proportion  also. 
A  double  sum,  to  run  the  same  time  and  at  the  same 
rate,  will  require  a  double  discount,  &c.  &c.  Now,  if 
we  assume  a  principal,  and  compute  the  amount  on  it 
for  the  given  rate  and  time,  that  amount  will  be  to  its 
principal,  as  the  given  amount  to  its  required  principal. 
If  we  take  unity  for  the  assumed  principal,  and  compute 
the  amount  corresponding  to  the  given  rate  and  time  giv- 
en in  any  problem,  that  amount  will  be  the  1st  term  of 
a  proportion,  one  (or  unity)  the  2nd  term,  and  the  given 
sum  the  ,3rd  term.  But,  as  the  2nd  term  is  one,  we 
find  the  4th  term  by  dividing  the  3rd  term  by  the  1st. 
Therefore,  to  obtain  the  present  worth  of  a  future  pay- 
ment, we  have  the  following 

Rule.  Divide  the  given  sum  by  the  amount  of  one 
dollar,  or  one  pound,  for  the  given  rate  and  time.  The 
difference  between  the  given  sum  and  present  worth,  is 
the  discount. 

EXAMPLES. 

].  What  is  the  present  luorth  and  discount  on  $150, 
due  9  months  hence,  discounting  at  the  rate  of  6  percent.? 


172 


ARITH3IETIC. 


The  amount  of  1  dollar  for  9  months,  at  6  per  cent.,  is 
1,045;  hence,  as  1,045  :  1  :  :  150  :  present  worth. 
By  rule : 

1,045)150,000(143,53  -j- 
1045 


4550 
4180 

Given  sum,  . 
Present  value 

Discount, .  .  . 

3700 
3135 

.  143,53=Prin. 

.  $6,47 

5650 
5225 

3250 
3135 

115 
2.  A  holds  a  note  of  375  dollars,  due  in  90  days,  but 
is  much  in  want  of  ready  money,  which  the  debtor 
agrees  to  pay,  provided  A  will  allow  discount  at  7  per 
cent.;  what  would  be  the  discount?  Amount  of  $1,  3 
months,  at  7  per  cent.,  is  1,0175.     By  rule, 

1,0175)375,0000(368,55  + 
30525 


69750 
61050 

87000 
81400 


Given  sum,  •  •  •  375= Amount. 
Present  worth,   368,55 =Prin. 


Discount, 


.  $6.45,  Jns. 


50000 

50875 

51250 

50875 


3750 


DISCOUNT    AND    EANKIXG.  173 


Let  the  pupils  notice,  that  the  sums  we  call  present 
values,  in  the  precedinof  examples,  are  principals,  which, 
put  at  interest  for  their  respective  rates  and  times,  would 
give  the  amounts  stated  in  the  questions. 

3.  A  person  pays  a  debt  of  394  dollars,  1  year  and  8 
months  before  it  is  due,  discounting  at  0  per  cent.;  what 
must  he  pay?  Ans.  $358,18  -h 

4.  What  is  the  present  worth  of  775  dollars  50  cents, 
due  in  4  years,  at  5  per  cent,  per  annum? 

Jln.'i.  t|;564G,25. 

5.  What  is  the  present  worth  of  580  dollars,  due  in  8 
months,  at  6  per  cent,  per  annum?     Jhi^.  $557,69  -\- 

6.  What  is  the  present  worth  of  954  dollars,  due  in  3 
years,  at  41,  per  cent,  per  annum  .'    .uUh.  $840,528  + 

7.  What  is  the  discount  of  205  dollars,  due  in  15 
months,  at  7  per  cent,  per  annum  ?     .^ns.  $16,495  + 

8.  Bought  goods  amounting  to  775  dollars,  at  9  months' 
credit;  how  much  ready  money  must  be  paid,  allowing 
a  discount  of  5  per  cent,  per  annum  ?    Ans.  $746,987. 

9.  B  owes  A  to  the  value  of  1005  dollars,  to  pay  as  fol- 
lows: viz.  475  dollars  in  10  months,  and  the  remainder 
in  15  months;  what  is  the  present  wortli,  allovv^ing  dis- 
count at  6  per  cent,  per  annum  ?  Ans.  $945,404. 

10.  What  is  the  difference  between  the  interest  of  2260 
dollars,  at  6  per  cent,  per  annum,  for  5  years,  and  the 
discount  of  the  same  sum,  for  the  same  time  and  rate  per 
cent.?  Ans.  $156,462  + 

(Art.  99.)  Discount  and  Bank  interest  ought  to  be 
the  same,  and  are  the  same  in  the  state  of  New  York; 
but  elsewhere,  even  in  England,  as  far  as  the  author's  in- 
formation extends,  bank  interest  is  simple  interest,  on 
the  face  of  a  note  paid  in  advance  ;  not  interest  on  the 
present  worth — not  interest  on  what  the  drawer  actually 
receives — but  interest  on  what  he  promises  to  pay  at  a 
specified  time.  For  example,  A  presents  his  note  at  a 
bank,  properly  endorsed  to  run  a  year,  the  legal  interest 
of  the  state  being  6  per  cent.;  the  bank  gives  him  $94, 
reserving  6  dollars  for  discoxmt.  Now,  94  dollars,  put 
out  at  interest  for  a  year,  at  6  per  cent.,  will  not  amount 

to  100  dollars;  therefore,  the  rate  per  cent,  that  a  bank 
_ 


174  ARITHMETIC. 


draws  interest,  is  more  than  the  nominal  rate — in  fact, 
more  than  lawful  interest.-  Banks  rarely  discount  paper 
that  has  more  than  4  months  to  run ;  they  always  prefer 
short  periods,  for  two  reasons.  One  is,  that  it  is  difficult 
to  look  far  into  futurity,  and  decide  on  future  circumstan- 
ces;  and  the  other  is!^  that  short  intervals,  and  advance 
payments  of  interest,  give  them  all  the  advantages  of 
compound  interest.  Bank  notes — or  rather,  notes  dis- 
counted by  a  bank — run  three  days  longer  than  tJie  time 
specilied,  called  three  days  of  grace;  but  the  interest  is 
computed  for  the  three  days,  and  the  note  need  not  be 
taken  up  until  the  last  day. 

From  the  preceding  explanations,  it  is  evident  that  Bank 
interest  must  be  compu  ^d  as  simple  interest,  on  the  face 
of  the  note  adding  three  days  to  llie  time — except  in  New 
York,  where  bank  interest  is  discount,  as  computed  by 
the  rule  under  Art.  98. 

EXAMPLES. 

1 .  A  note  for  350  dollars,  for  90  days,  was  discounted  at 
bank  ;  what  was  the  discount  ? 

Operation. 
Rule,  under  Art.  94,  .  .  •  20  /t0(^  1  3^^0     7 
2     ^0  1    ^^  31 

4,0)217 


.fins,'  •  .$5,42',. 

2.  What  is  the  bank  interest  on  a  note  of  135  dollars, 
payable  90  days  after  date?  Ans.  $2,091  -f 

3.  What  is"^the  bank  interest  on  a  note  of  1850  dol- 
lars, payable  90  days  after  date?  Ans.  $28,671. 

4.  What  is  the  bank  interest  on  a  note  of  475  dollars, 
payable  30  days  after  date  ?  Ans.  82,61. 

5.  AVhat  will  be  the  proceeds  of  a  note  of  1370  dol- 
lars, payable  30  days  after  date  ?  Ans.  1362,46|. 


*  Yet,  the  supreme  courts  ha  '3  decided,  that  this  mode  of  cal- 
culating discount  is  not  usury  Banks  are,  therefore,  authorized 
by  judicial  authority  to  discount  at  a  higher  rate  than  that  establish- 
ed by  law.  We  know  not  on  what  principles  this  decision  has  been 
made. 


DISCvOUNT    AND    BANKING.  175 


6.  What  will  be  the  proceeds  of  a  note  of  880  dollars, 
payable  90  days  after  date  ?  ^2ns.  $860,3G. 

MISCELLANEOUS    PROBLEMS    IN    INTEREST. 

(Art.  100.)  The  following  problems  are  best  nnder- 
stood  by  cause  and  effect,  as  we  have  already  explained 
(Art.  84).  We  only  give  a  few  more  examples  in  this 
place,  as  being  in  local  order. 

1.  A  loaned  1600  dollars  at  6  per  cent.,  until  it  amount- 
ed to  2000  dollars  ;  what  was  the  time  ? 

Ans.  4  years  2  montlis. 
Cause.     Effect.       Cause.     Effect. 
100      :      6     : :      1600     :     400 
Statement,  •  •  •     1.2  fl 

2.  A  sum  of  money  was  put  out  at  interest,  at  7  per 
cent.,  and  remained  3  years  5  months  and  10  days,  and 
the  interest  was  $15,57;  what  was  the  principal  ? 

Ans.  $64,58. 

Cause.     Effect.       Cause.     Effect. 

Q,  ,  ,  100      :      7       ::       []    :     1^,57 

Statement,  ...      ,^  k-[ 

3.  At  what  rate  per  cent,  per  annum,  will  1200  dollars 
amount  to  1476  dollars,  in  5  years  and  9  months? 

Ans.  4  per  cent. 

4.  If  834  dollars,  at  interest  2  years  and  6  mouths, 
amount  to  927  dollars  82^  cents,  what  was  the  rats  per 
cent,  per  annum  ?  Jlns.  4,1^  per  cent. 

5.  Sixty-three  dollars  was  at  interest  1  year  3  mouths, 
and  yielded  4  dollars  72^  cents;  what  was  the  rate? 

.Sns.  C. 

6.  The  interest  on  a  certain  sum,  for  6  years  and  3 
months,  at  6  per  cent.,  was  28|-  dollars ;  what  was  the 
sum?  Ans.  $75. 

7.  The  interest  on  730  dollars,  at  6  per  cent.,  was  253 
dollars  31  cents;  what  was  the  time? 

Ans.  5y.  9  m.  12  d. 

8.  Six  hundred  and  thirty  dollars  was  put  at  interest, 
at  6  per  cent.,  and  the  interest  received  was  25  dollars 
20  cents;  what  was  the  time  out?        Ans.  8  months. 

9.  Seven  hundred  and  fifty  dollars  was  at  interest  9 


170  ARITHMETIC. 


months,  and  the  amount  received  was  $789,375  ;  what 
was  the  rate  per  cent?  ^8ns.  7. 

10.  If  300  doUars  gives  45  dollars  interest  in  5  years, 
what  was  the  rate  per  cent.?  Ans.  3. 

11.  The  amount  of  750  dollars,  at  6  per  cent.,  was 
1301  dollars  25  cents ;  what  time  was  it  out? 

Ans.  12  y.  3  m. 

12.  The  interest  on  a  certain  svim  at  4  per  cent.,  for  4 
months  and  18  days,  was  9  dollars  20  cents  ;  what  was 
the  sum  ?  Ans.  $600. 

13.  In  what  time  will  837  dollars  amount  to  1029  dol- 
lars 51  cents,  at  5 J  per  cent.?  .dns.  4  years. 

14.  What  sum  will  amount  to  571  dollars  20  cents  in  4 
years,  at  5  per  cent.?     [This  comes  under  discount.] 

Ans.  $476. 

15.  The  interest  on  576  dollars,  at  6  per  cent.,  is 
$60,48;  what  was  the  time?  Ans.  21  months. 


PERCENTAGE. 


(Art.  101.)  This  indxdes  Commission,  Broherage  and 
Insurance;  all  of  which  is  paid  at  so  much  per  hundred; 
hence,  it  is  called  per-centage,  and  must  fall  back  on  pro- 
portion ;  or,  it  may  be  more  naturally  compared  to  in- 
terest, wanting  the  element  of  time,  or,  as  in  the  case  of 
insurance,  considering  no  other  than  the  iinit  of  time ; 
but  commission  and  brokerage  have  no  reference  to  time 
whatever;  we  only  consider  the  proportion  to  the  given 
rate  per  100 :  hence  the  following 

Rule.  Jis  100  is  to  the  given  rate  per  cent.,  so  is  the 
given  sum  to  the  required  commission,  brokerage,  or 
insurance.'-     Or, proceed  in  the  same  manner  as  though 


■*  Commission  is  compensation  for  selling  or  buying  goods  for  eth- 
ers. Brokerage  is  allowance  made  to  dealers  in  money  or  stocks, 
for  the  purchase  of  stock  or  the  exchange  of  money,  or  commercial 
paper.  Insurance  is  money  paid  to  a  joint-stock  company,  for  them 
to  assume  the  hazard  and  risk  of  damage  by  fire,  or  loss  of  any  kind 
by  sea;  and,  in  case  of  damage  or  loss,  the  company  or  insurers 
are  to  make  such  loss  good,  as  expressed  in  the  contract.     The  written 


PER-CENTAGE.  177 


you  were  required  to  find  the  interest  of  the  given  sum 
for  one  year. 

EXAMPLES. 

1.  An  agent  sold  goods  for  his  employer  to  the  amount 
of  1200  dollars  ;  what  was  his  conmiission  at  5  per  cent.-? 
Statement:  as  100  :   5  ::  1200  :  to  answer.  J]ns.  $60. 

2.  Sold  goods  to  the  amonnt  of  975  dollars  50  cents, 
on  a  commission  of  7^^  per  cent.;  what  was  niv  compen- 
sation? "  /]ns.  $73, J 6^. 

3.  A  merchant  exchanges  3250  dollars  of  unbankable 
notes  with  a  broker,  allowing  him  1^^^  per  cent,  for  par 
funds  ;  how  much  did  the  broker  pay  him  ? 

By  practice  ;  thus,  at  1  per  cent.,  the  broker's  bonus 

or  commission  would  be, $32,50 

At  i  per  cent., 16,25 

1    "      "        8,125 


Broker's  full  commission, $56,875 

Therefore,  the  merchant  must  receive  3193  dollars  12^ 
cents,  the  remaining  part  of  the  3250  dollars. 

4.  An  agent  sells  750  bales  of  cotton,  at  48  dollars  per 
bale,  and  received  H  per  cent.  coRimission ;  how  much 
did  he  receive? 

750 


Q  1   ,.  100 

Solution : 

2 


48 
3  .^ns.  $540. 

5.  What  must  be  paid  to  insure  a  steamboat  and  cargo, 
from  Pittsburgh  to  New  Orleans,  at  a  premium  of  |  of  1 
per  cent.,  the  valuation  of  the  whole  being  47500  dollars? 

/Jns.  $356,25. 

6.  A  gentleman  in  New  York  has  a  house  valued  at 
15000  dollars ;  he  can  obtain  insurance  lor  12000  dollars, 
at  a  premium  of  2|  per  cent.:   what  is  his  insurance  tax  ? 

.^ns.  $285. 

7.  The  sales  of  certain  goods  amount  to  1680  dollars; 
what  sum  is  to  be  received  for  them,  allowing  2^  per 
cent,  for  commission?  ^^ns.  $1633,80. 


instrument  expressing  such  contract,  is  called  a  policy      The  amount 
of  insurance  paid,  is  called  the  premium. 


178  ARITHMETIC. 


8.  What  is  the  insurance  of  760  dollars,  at  6!,-  per 
cent.?  ^'liis.  $49,40. 

9.  What  is  the  insurance  of  5630  dollars,  at  7^  per 
cent.1  ^ns.  436,325. 

10.  A  merchant  sent  a  ship  and  cargo  to  sea,  in  time  of 
war,  valued  at  17654  dollars;  what  would  be  the  amount 
of  insurance,  at  18}  per  cent.?  .^ns.  $3310,12',. 

11.  What  is  the  brokerage  on  21'50  dollars,  at  2  per 
cent.?  ^^ns.  $43. 

12.  A  merchant  shipped,  on  board  a  vessel  bound 
from  Boston  to  St.  Thomas,  goods  to  the  amount  of  2464 
dollars ;  what  must  he  pay  to  be  insured  against  the  dan- 
gers of  the  sea,  the  premium  being  2j  per  cent.? 

^)is.  $55,44. 

13.  A  merchant  having  a  house,  valued  at  10650  dol- 
lars, and  goods  in  his  store  amounting  to  6740  dollars ; 
what  must  he  pay  to  be  insured  against  the  danger  of  tire, 
the  premium  being  U  per  cent.?  ^ns.  $260,85. 

14.  A  sends  4500  dollars  to  an  agent,  to  purchase  rail- 
road stock,  and  allows  the  agent  2  per  cent,  for  his  trou- 
ble in  making  the  investment ;  how  much  of  the  money 
shall  the  agent  retain?  .^ns.  $88,23  -f 

(Art.  102.)  In  this  case,  it  is  evident  that  the  agent 
must  not  compute  per-centage  on  that  portion  of  the  mo- 
ney which  he  may  retain.  It  is  only  2  per  cent,  on  the 
portion  spent  for  his  employer.  For  every  100  dollars 
the  agent  spends,  he  is  to  have  2  dollars ;  then  for  every 
102  dollars,  100  is  to  go  to  the  principal,  and  the  remain- 
der to  the  agent.  Therefore,  as  102  :  100  :  :  4500  : 
a  4th  term.  This  4th  term  is  $4411,76  4-?  spent  for  the 
employer,  and  the  remaining  $88,23  -{-,  is  retained  by  the 
agent.     Strictly  speaking,  this  is  diaeount. 

(Art.  103.)  Neither  individuals  nor  insurance  compa- 
nies will  ever  insure  to  the  full  value  of  property,  as  this 
might  create  negligence  in  those  who  have  care  of  it;  or 
worse,  tempt  the  unprincipled  to  destroy,  or  suffer  its  de- 
struction, to  turn  it  into  cash.  Yet,  in  tiie  various  turns 
of  business,  it  is  sometimes  necessary  for  individuals  to 
secure  a  policy  for  a  given  specified  amount^  even  at  the 
expense  of  paying  a  higher  rate.     Such  a  policy  may  be 


STOCKS.  179 

wanted,  to  send  before  a  vcsse'  to  a  foreign  port,  lo  sus- 
tain the  credit  of  a  partner,  agent,  otr  friend,  until  the  in- 
sured siiip  or  cargo  sliall  arrive. 

To  secure  a  given  amount,  in  case  of  tlie  destruction 
of  the  property,  we  must  draw  for  a  higher  sum,  such  a 
sum  as  when  the  per-centage  is  taken  off,  will  leave  the 
given  amount. 

Now,  as  all  these  operations  are  based  on  proportion, 
it  is  evident  that  we  must  take  the  per-centage  from  100: 
then  say, 

,fis  the  remainder  is  to  100,  no  is  the  desired  amount 
to  the  policy  to  be  taken  out. 

Example  1.  A  merchant  wishes  to  secure  $1920  on 
an  adventure  to  London  ;  what  sum  must  be  mentioned 
in  the  policy  to  cover  that  amount,  the  premium  being  4 
per  cent.? 

As  96  :   100  ::   1920  :  fourth  term  ; 

or,     1   :   100  ::  20  :  2000.       Jins.  $2000. 

2.  A  gentleman  has  produce  on  hand  to  the  amount  of 
$14000,  which  he  intends  to  ship  to  Soutli  America. 
The  marine  insurance  compaay  are  willing  to  insure  to 
the  amount  of  $9000  for  2\  per  cent.  ;  but  he  wishes 
to  secure  $11000,  and  for  this  sum  they  demand  a  pre- 
mium of  2|  per  cent.;  what  sum  must  be  mentioned  in 
the  policy  ?  Ans.  $  1 1 3 1 1 ,05. 

3.  A  lady  has  property  on  which  she  wishes  to  secure 
the  sum  of  $15000;  what  must  be  the  policy  to  cover 
that  sum,  the  premium  being  6^  per  cent.? 

Ans,  $16042,79. 


STOCKS. 


(Art.  104.)  Stock  is  the  term  given  to  the  aggregate 
or  any  number  of  shares  of  the  capital  belonging  to  any 
legal  association  or  company,  created  for  any  special  pur- 
pose, such  as  banking,  insurance,  making  public  improve- 
ments, or  establishing  manufactories. 

To  organize  a  bank,  such  as  the  people  will  have  con- 


180  ARITHMETIC. 


fidence  in,  it  is  necessary  that  it  should  have  ready  capi- 
tal at  all  times,  to  redeem  its  notes  ;  and  to  secure  this 
end,  a  certain  amount  of  capital  is  designated  in  the  char- 
ter or  legal  act  of  corporation  :  this  capital  is  divided  into 
shares,  H^enerally  consisting  of  $100  each,  for  which  indi- 
viduals subscribe  and  pay.  These  individuals  are  called 
stockholders,  and  receive  the  profits  of  the  bank,  in  pro- 
portion to  the  shares  they  hold.  Certificates  of  stock  are 
bought  and  sold  like  individual  notes  or  other  property ; 
and  when  the  profits  of  the  company  are  large,  the  stock 
will  command  more  than  its  nominal  value  in  the  mar- 
ket, and  is  then  said  to  be  above  par  ;  and  when  the  pro- 
fits are  small,  the  stock  sinks  below  par. 

The  rates  above  and  below  par  are  estimated  at  so 
much  on,  or  so  much  taken  ofT  of  the  100.  Hence,  we 
compute  the  value  of  stocks  by  per-centage,  as  in  Art. 
101,  or  Art.  103. 


EXAMPLES. 


1.  A  sells  25  shares  of  bank  stock  at  3  per  cent,  ad- 
vance, the  par  value  per  share  being  $100  ;  how  much 
money  did  he  receive?  ^^ns.  $2575. 

Statement :    If  100  give  103,  what  will  2500  give  ? 

2.  The  stock  in  a  certain  rail-road  company  is  15  per 
cent,  below  par;  what  are  12  shares  worth,  the  original 
value  being  $50  per  share  ?  ^ns.  $510. 

Statement:   100  :  85  :  :   12X50  :  .^ns. 

3.  At  a  time  of  great  fluctuation  in  stocks,  a  gentleman 
bought  36  shares  of  the  United  States  Bank  stock,  nom- 
inal value  $100,  at  17|-  per  cent,  advance,  and  sold  the 
same  day  at  181^  advance;  what  did  he  realize  by  the 
transaction  ? 

^7is.  His  gain  was  |  of  1  per  cent.,  or  $27. 

4.  If  the  stock  of  an  insurance  company  sells  for  3i 
per  cent,  below  par,  what  are  $1200  of  the  stock  worth  ? 

Ans.  $1159,50. 

5.  I  directed  a  broker  to  purchase  for  me  75  shares  in 
a  certain  rail-road  company,  if  he  could  obtain  them  at 


EQUATION    OF    PAYMENTS.  181 


22  per  cent  below  par,  promising  him  f  per  cent,  for  his 
trouble  ;  what  did  the  whole  cost  me  ? 

^ns.  $5886,56  -}- 

[Note.  "When  the  original  or  nominal  value  of  a  share 
is  not  mentioned,  100  is  understood.] 

6.  Bought  56  shares  in  a  New  York  state  bank,  at  7/, 
per  cent,  advance,  the  original  value  being  $75  per  share  ; 
what  will  the  whole  cost  me,  allowing  |  per  cent,  to  the 
broker,  who  made  the  purchase  ?  Ans.  $4530,75. 


EQUATION  OF  PAYMENTS. 

(Art.  105.)  By  this  is  meant  the  equitable  time  of 
paying  several  debts  due  at  ditferent  times,  so  that  neither 
party  siiall  sustain  any  loss  of  interest. 

To  explain  more  fully,  let  us  suppose  that  a  person 
owes  his  neighbor  as  follows  : 

2  dollars,  to  be  paid  in  2  months  ; 
2  dollars,  to  be  paid  in  3  months ;  and 
2  dollars,  to  be  paid  in  5  months. 
In  what  time  shall  he  pay  the  whole  6  dollars,  so  that 
neither  himself  nor  creditor  shall  lose  interest?     We  an- 
alyze it  thus  : 

2  dollars  in  2  months  will  gain  as  much  interest  as 

4  dollars  in  1  month  ; 
2  dollars  in  3  months  will  gain  as  much  interest  as 

6  dollars  in  1  month  ; 
2  dollars  in  3  months  will  gain  as  much  interest  as 
10  dollars  in  1  month.     Then 

$6  for  the  several  times  will  g'ain  as  much  interest  as 
$20  for  1  month. 
But  there  are  only  6  dollars  to  be  paid,  not  20  ;  and  as 
interest  is  always  in  proportion  to  the  compound  of  mo- 
ney and  time,  we  must  find  ivhat  number  of  months, 


182  ARITHMETIC. 


multiplied  by  G,  will  give  the  same  product  ari  20,  multi- 
plied by  1  ;  and  that  evidently  is  -^'  =3^-  months. 
From  this  we  may  derive  the  following  general 

Rule.  Mulliply  each  payment  by  its  thne,  and  di- 
vide the  sum  of  the  several  products  by  the  sum  of  the 
jjayments,  and  the  quotient  will  be  the  equated  time  for 
the  payment  of  the  whole, 

EXAMPLES. 

1.  If  600  dollars  are  now  due,  600  dollars  in  4  months, 
and  600  dollars  in  8  months,  600  dollars  in  12  months, 
what  is  the  equitable  time  for  paying  the  whole  ? 

N.  B.  It  is  not  necessary  in  this  or  any  other  prob- 
lem, that  we  should  use  the  sums  of  money  actually  giv- 
en. It  is  sufficient  that  Ave  use  the  same  proportional 
parts.  In  this  case  we  will  use  one  dollar  in  place  of  600, 
and  the  operation  will  stand  thus : 

IX  0=  0 
IX  4=  4 
IX  8=  8 
1X12=12 


4)        =24(6  months,  ,^ns. 

2.  If  750  dollars  are  to  be  paid,  |  of  it  in  H  years, 
~  of  it  in  2  years,  and  the  residue  in  2^  years,  what  is 
the  equated  time  of  paying  the  whole  at  once  ?    • 

./???5.  23  J  months. 
In  this  example,  it  is  not  necessary  to  use  750  dollars; 
we  had  better  use  10. 

3.  A  owes  B  100  dollars,  to  be  paid  in  6  months  ;  120 
dollars,  to  be  paid  in  10  months,  and  160,  to  be  paid  in 
14  months;  what  is  the  equated  time  for  paying  the 
whole?  .^ns.  10|^r  months. 

4.  A  merchant  haUi  owing  to  him  300  dollars,  to  be 
paid  as  follows:  50  dollars  at  2  months,  160  dollars  at  5 
months,  and  the  residue  at  8  months;  and  it  is  agreed  to 
make  one  payment  of  the  whole  ;  when  must  that  time 
be?  Jlns.  6  months. 

5.  F  owes  H  2400  dollars,  of  which  480  dollars  are 


EQUATION    OF    PAYMENTS.  183 


lo  be  paid  present,  960  dollars  at  5  months,  and  the  rest 
at  10  months;  but  they  agree  to  make  one  payment  of 
the  whole,  and  wish  to  know  the  time? 

A.rins.  6  montlis. 

6.  A  merchant  bought  goods  to  the  amount  of  2000 
dollars,  and  agreed  to  pay  400  dollars  at  the  time  of  pur- 
chase, 800  dollars  at  5  months,  and  the  rest  at  10  months  ; 
but  it  is  agreed  to  make  one  payment  of  the  whole ;  what 
is  the  mean  or  equated  time  I  Ans.  G  months. 

N.  B.  To  solve  the  following,  we  had  better  fall  back 
on  the  general  principle,  and  not  attempt  to  apply  the 
rule. 

7.  A  owes  B  600  dollars,  to  be  paid  in  2  years  from 
the  date  of  the  note ;  but,  at  the  expiration  of  6  months, 
A  agrees  to  pay  150  dollars,  if  B  will  wait  enough  long- 
er for  the  balance,  to  compensate  for  the  advance ;  how 
long  ought  B  to  wait? 

Ans.  6  months  after  the  2  years. 

Here  150  dollars  was  paid  18  months  before  the  time  ; 
how  long  shall  the  remaining  450  dollars  remain  after  the 
expiration  of  the  2  years,  to  give  the  same  interest. 

Statement:   150X18  :  450X[]  :  :   1   :   1. 

m 

8.  P  owes  Q  420  dollars,  which  will  be  due  6  months 
hence  ;  but  P  is  willing  to  pay  him  60  dollars  now,  pro- 
vided he  can  have  the  rest  forborne  a  longer  time  ;  it  is 
agreed  on;  tlie  time  of  forbearance,  therefore,  is  required? 
Ans.  7  months  ;  that  is,  1  month  in  addition  to  the  6. 

9.  A  young  gentleman  has  a  legacy  of  1500  dollars, 
to  be  paid  him  in  16  months  ;  but,  being  in  want  of  ready 
money,  agrees  to  defer  the  payment  of  the  balance  the 
proper  time,  if  he  can  have  500  dollars  in  hand  ;  when 
shall  the  balance  be  paid  ?  Ans.  2  years  hence. 


See  Art.  80.  ^  ^^0 


184  ARITHMETIC. 


PROFIT  AND  LOSS  PER  CENT. 

(Art.  106.)  When  articles  are  bought  and  sold,  it  is 
often  desirable  to  know  the  rate  of  gain  or  loss  per 
cent.,  corresponding  to  any  definite  or  assumed  price. 
This  of  course  can  be  done  hy  proportion,  Practice,  and 
proportion.  Interest  and  discount  are  all  the  arithmeti- 
cal principles  that  can  be  brought  into  requisition,  under 
this  head ;  and  practice,  interest,  and  discount,  are  all  re- 
solvable into  joro/)or/20?i;  hence  proportion  alone  is  the 
sole  i^rinciple, 

EXAMPI.ES. 

1.  A  merchant  bought  cotton  cloth  at  1  shilling  6 
pence  per  yard,  and  sold  it  at  1  shilling  10  pence;  what 
was  his  gain  per  cent.? 

pence,  pence,  gain. 
Statement:  If  18  :  4  : :   100  :  how  much? 

Ans.  22f . 

2.  A  grocer  bought  tea  at  60  cents  per  pound,  and  sold 
the  same  at  75  cents;  what  was  his  profit  per  cent.? 

Ans.  25. 

As    60      :     15     ::     100     :    Answer 
Statement:    ^.^      4      .        1     : :      100     :       25 

3.  I  bought  Irish  linen  at  56  cents  per  yard ;  what  shall 
I  sell  it  for,  to  gain  20  per  cent.?  Ans.  67 fo- 

Statement:     As   100  :   120  : :  56  :    Answer. 

4.  A  merchant  sold  broadcloth  at  4  dollars  75  cents 
per  yard,  making  a  profit  of  32  per  cent.;  what  did  the 
cloth  cost  him  ?  Ans.  $3,60,  nearly. 

Statement:  132,  that  he  now  receives,  originally  cost 
him  100  ;  in  that  proportion,  what  did  475  cost?  or,  as 
132  :    100  ::   475  :  Answer. 

5.  A  merchant  bought  English  broadcloth  ;  the  first 
cost,  duties  and  transportation,  amount  to  14  shillings  8 
pence  per  yard  ;  what  must  be  his  price,  in  dollars  and 
cents,  to  gain  20  per  cent.,  the  dollar  being  equal  to  4 
shillings  6  pence  sterling?  Ans.  $3,91  -|- 


PROFIT    AND    LOSS     PER    CENT.  185 


Statement:  100  :  120  ::  14|  :  price  in  shillings  sterl- 
ing. But  to  bring  shillings,  sterling  money,  into  dollars, 
we  must  divide  by  4\,  or  multiply  by  2  and  divide  by  9. 
Tiie  whole  combined  in  one  c  peration  (Art.  24),  stands 
thus  : 


100 
9 


44 

^0  4 
2 


6.  If  I  purchase  Irish  linen  at  2  shillings  4  pence  per 
yard,  sterling  money,  what  must  I  sell  it  at  per  yard,  in 
federal  money,  to  gain  30  per  cent.?  ^^ns.  67^4. 

7.  A  man  bought  300  sheep,  at  2  dollars  15  cents  per 
head,  and  his  expense  in  making  the  purchase,  was  45 
dollars  50  cents ;  he  sold  them  at  3  dollars  20  cents  per 
head;  what  was  his  whole  gain,  and  what  was  his  gain 
percent.?  ^ns.  Whole  gain,  $269,50;  gain  per  cent. 
39,  nearly. 

8.  If  I  buy  12'2  hundred  weight  of  sugar  for  140  dol- 
lars, how  much  must  it  sell  at  per  pound,  to  make  25  per 
cent.?  ^ns.  12^-  cents. 

Solved  as  in  Practice  (Art.  88), 

>  cents. 


140 
100 

100 

125 

CWf.' 

.  12L 

qr.    .  . 

.   4 

. 

lb..   . 

.  28 

This  cancels  down  to  the  answer. 

9.  Bought  126  gallons  of  wine  for  150  dollars,  and 
retailed  it  at  20  cents  per  pint ;  what  was  the  whole  gain, 
and  what  the  gain  per  cent.? 

Ans.  Whole  gain,  $51,60;  gain  per  cent.  34|. 

10.  Bought  a  hogshead  of  molasses  for  26  cents  per 
gallon,  and  suppose  it  lost  5  per  cent,  in  waste  and  leak- 
age ;  how  must  I  sell  the  remainder,  per  gallon,  to  gain 
25  per  cent.?  Ans.  34yV. 

11.  If,  by  selling  1  pound  of  pepper  for  10^  cents, 
there  are  %  cents  lost,  how  much  is  the  loss  per  cent.? 

Ans,  16. 

J2 


186  ARITHMETIC. 


12.  A  merchant  bought  180  casks  of  raisins  at  2  dol- 
lars 13  cents  per  cask,  and  sells  them  at  3  dollars  68 
cents  per  hundred  iveif(ht,  and  gains  25  per  cent.;  re- 
quired the  average  weight  in  each  cask. 

jlns.  8 1/2  pounds. 
cost.  sale. 

Solution:    180X213    :    368X[]    ::    100    :    125 
A  proportion  wanting  a  factor  to  one  term,  that  factor  is 
the  number  of  hundred  weis^ht,  Avhich  must  be  reduced 
to  pounds,  and  divided  by  180.     By  Art.  24  :  thus, 
368     ^^0 
4  ^00     213 

;t^0     ^^^   ^ 
112 

13.  A  merchant  receives  raisins,  which  cost  2  dollars 
50  cents  per  cask,  and,  by  selling  them  at  5  cents  a  pound, 
lie  gains  20  per  cent,  on  the  first  cost ;  what  was  the  av- 
erage weight  of  each  cask  ?*  ^^ns.  60  lbs. 

14.  What  is  the  loss,  and  what  is  the  loss  per  cent., 
in  laying  out  70  dollars  for  hats,  at  1  dollar  75  cents  each, 
and  selling  them  at  25  cents  a-piece  less  than  cost? 

Jns.  Whole  loss,  $10,  loss  per  cent.  14f . 

15.  A  merchant  bought  1200  pounds  of  coffee  for  135 
dollars  ;  what  must  he  sell  it  at  per  pound,  to  gain  35 
dollars  on  the  whole?  »^ns.  14^  cents. 

16.  Bought  2  hogsheads  of  wine,  at  1  dollar  25  cents 
per  gallon,  and  sold  the  same  at  1  dollar  60  cents  ;  what 
was  the  whole  gain,  and  the  gain  per  cent.? 

.^ns.-  Whole  gain,  $44,10;  gain  per  cent.  28. 

17.  If,  by  selling  cloth  at  4  dollars  50  cents  a  yard,  I 
lose  20  per  cent.,  what  was  the  prime  cost? 

.^715.  $5,621. 

18.  Sold  corn  at  25  cents  per  bushel, and  lost  3  cents: 
what  was  the  loss  per  cent.?  ^ins.  10+. 

(Art.  107.)  When  persons  sell  on  credit,  it  is  evident 
that  discount  must  be  taken  off,  before  we  have  the  true 


*  This  problem  is  extracted.  Its  author  mentions  200  casks;  we 
omit  the  number,  as  it  will  cancel  itself,  whatever  it  be,  as  seen  in  the 
preceding  problem. 


PROFIT  AND  LOSS  PER  CENT.  187 


profit;  also,  when  goods  have  been  on  Imnd  some  time, 
the  interest  of  tirst  cost  shoukl  be  added  on,  before  we 
really  have  the  prime  cost. 

19.  Sold  goods  to  the  amount  of  425  dollars,  on  6 
months  credit,  which  was  25  dollars  more  than  the  goods 
cost  me  ;  what  was  the  true  profit,  money  being  worth  G 
per  cent.?  JJns.  $12,62. 

20.  A  merchant  sold  goods  for  667  dollars,  to  be  paid 
in  8  months  :  the  same  goods,  one  year  ago,  cost  him 
600  dollars,  apparently  gaining  67  dollars ;  what  was  his 
true  gain,  money  commanding  6  per  cent.? 

An8.  $5,346  -}- 
[Note.  That  is,  he  gained  5  dollars  and  .34  cents,  on 
the  supposition  that  he  traded  on  borrowed  capital,  or 
on  the  supposition  that  mterest  is  7iot  gain.  But  this  is 
not  true  ;  interest  is  gain,  and  legal  interest  is  as  much 
as  mere  capital  ought  to  gain.  In  the  case  under  consid- 
eration, the  merchant  gained  5  dollars  and  34  cents  in 
trading  in  goods,  more  than  it  is  supposed  he  could  have 
done  by  trading  in,  or  loaning  money.] 

21.  A  merchant  has  had  a  certain  lot  of  goods  on  his 
hands  6  months,  and  now  has  an  opportunity  to  sell  on 
10  months  credit,  at  an  advance  of  30  per  cent,  on  the 
original  cost ;  what  are  his  profits  per  cent.,  when  he  is 
paying  5  per  cent,  interest  on  capital  ? 

.%is.  21i 

22.  Sold  goods  to  the  amount  of  723  dollars,  ready 
cash,  at  15  per  cent,  above  first  cost ;  what  was  my  real 
gain,  having  had  the  goods  on  hand  1  year  and  6  months, 
money  being  worth  6  per  cent.:  that  is,  how  much  more 
do  I  gain  by  the  purchase  and  sale  of  the  goods  than  I 
should  have  gaified  by  putting  the  original  cost  of  the 
goods  at  interest?  Jhis.  $37,73. 

23.  A  grocer  bought  3  barrels  of  sugar,  the  whole  nett 
weight  being  690  pounds,  at  8^^  cents  per  pound,  and 
sells  it  at  9'^  ;  what  is  his  whole  gain,  and  gain  per  cent.? 

.^ns  I  Wliolegain,  $10,35. 
'  5  Gain  per  cent.  IBfy. 

24.  A  grocer  bought  1'.  begs  of  coff'ee,  nelt  weight  of 
the  whole  being  1350  pounds,  a   12^  cents  per  pound; 


188  ARITHMETIC. 


what  must  be  his  price  per  pound,  to  gain   $50  on  tiie 
whole  ?  Ans.  10|-  cents. 

25.  Bought  a  cask  of  molasses,  containing  42  gallons; 
6  gallons  leaked  out,  and  I  sold  the  remainder  lor  37  \ 
cents  per  gallon,  and  gained  20  per  cent.  ;  what  was  the 
original  cost  per  gallon  I  Ans.  26^^  cents. 


REDUCTION  OF  CURRENCIES. 

(Art.  108.)  By  this  is  meant  the  changing  of  any  sum 
of  money  from  one  sfundard  to  its  equivalent  value  in 
another,  and  involves  no  other  principle  than  general  re- 
duction ;  and,  as  in  general  reduction,  the  operator  is 
guided  only  by  the  given  tabular  values. 

TABLE    OF    STATE    CURRENCIES. 

In  New  England  and  Virginia,  Kentucky  and 
Tennessee. 

3 
3 
18r/.  =     0,25  J       J  12^  cents=9  pence 

In  New  York,  Ohio  and  North  Carolina. 
2 
8 
24  d.  ■■=     ,25  (_      J  25  cents=2  shillings 

In  Pennsylvania,  N.  Jersey,  Delaware  and  Maryland. 

3  5.=     ,40 -j  or,  [.$1=7  5.  6  fZ. =90(7. 
9d.=     ,10  (        \  ,50=3  s.  9  d.=45  d. 


£=$10,00']       ■)Sl=,3i3 

.s.  =     0,50  lor,  I $1=6  shillings. 

d.=     0,25  J       J  12^  cents=9  pe: 

w  York,  Ohio  and  North  Carolin 
£=$5,00r      "]$1=,4£ 
5.  =$1,00 4  or,  [>$1=8  shillings. 
d.  ■■=     ,25  (_      J  25  cents=2  shil 

inia,  N.  Jersey,  Delaware  and  Jh 

£=$8,oor     ~)$i=-2je 

s.=     ,40ior,[$l=7s.6d.={ 
d.=     ,10  t      J  ,50=3  5.  9c/.= 

South  Carolina  and  Georgia. 
7^  =$30,00  ~)       •)U  =  ^\£ 
7  s.  =      1,50  lor,  I  $1=4  5.  8  ^Z.=i 
4  d.=       ,25  J       j    ,25  =  1  5.  2  c?. 


,      .  ,  =56ff. 

14 

[Note.  In  former  days,  before  congress  established 
our  national  currency,  all  accounts  were  kept  in  pounds, 
shillings  and  pence,  4  shillings  6  pence  corresponding  to 


REDUCTION  OF  CURRENCIES.  189 


the  Spanish  dolhir.  The  several  colonies,  as  these  states 
were  then  called,  issued  bills  of  credit,  and  passed  them 
as  money  ;  but  they  soon  depreciated,  and  moie  in  some 


tates  than  in  others,  and  at  the  general  adoption  of  fede- 
ral money,  the  values  of  the  pounds,  shilling,  Sic.  were 
left  as  noted  in  the  table.  These  currencies  are  legally 
dead,  and  ought  to  be  buried;  but  long  habit  becomes  se- 
cond nature  to  most  people,  and  the  detestable  spirit  of 
avarice  serves,  in  some  places,  to  keep  up  the  distinction. 
For  instance,  in  New  York,  where  the  penny  and  the 
cent  are  very  near  in  value  to  each  other,  the  petty  mer- 
chants and  shopkeepers  purchase  in  cents,  and  sell  in 
pence,  and  thereby  gain  4  per  cent. 

From  the  preceding  table,  to  change  pounds,  shillings, 
and  pence  into  dollars  and  cents,  we  have  the  following 

Rules. — 1.  Reduce  the  shillings  and  pence  lo  the  de- 
cimal of  a  pound  [Art.  69).  Then  state  the  probleyn  in 
proportion,  by  the  relation  of  £  and  $,  taken  from  the 
table. 

Or,  divide  the  given  number  of  £^s,  and  its  decimal, 
by  the  decimal  of  a  pound  that  makes  a  dollar. 

Or,  if  it  is  a  vulgar  fraction,  multiply  by  its  denom- 
inator, and  divide  by  the  numerator,  taking  care  to  can- 
cel, if  possible. 

2.  Reduce  the  given  pounds,  shillings  and  pence,  to 
shillings  and  decimals  of  a  shilling,  and  divide  by  the 
number  of  shillings  in  a  dollar;  or,  if  more  convenient, 
reduce  all  to  pence,  and  divide  by  the  number  of  pence 
in  a  dollar, 

examples. 

1.  Reduce  25  pounds  8  shillings  6  pence,  New  Eng- 
land currency,  to  dollars. 

Operation, 
6 


12 

20 
,3 


8,5 


25,425 =pounds  and  decimals  of  a  pound. 


Ans,  $84,75 


190 


ARITHMETIC. 


2.  Reduce  42  pounds  10  shillings  3  pence,  New  York 
currency,  to  dollars. 


1st  Operation. 


12 
20 
,4 


3 

10,25 

42,5125 


2nd  Operation. 
42  1  5.  3  c/. 
20 


8)850,25 


Ans.  $106,281 


Ans.  $106,281 
3.  Reduce  120  pounds  7  shillings  9  pence.  New  Jer- 
sey currency,  to  dollars. 

120     7     9 
20 


Is.  6rf.=7,5)2407,75($321,03-f 

4.  Reduce  38  pounds  5  shillings  6  pence,  Georgia  cur- 
rency, to  dollars. 
12     6 


20 


5,5 


£38,275.       Di\dding   by  /„,  gives  $161,178+ 

5.  In  11  shillings  6  pence,  Georgia  currency,  how  ma- 
ny dollars  and  cents  ?  Am.  $2,464+ 

6.  In  iElOO,  New  York  currency,  how  many  dollars  ? 

Ans.  $250. 

7.  In  £100,  New  England  currency,  how  many  dol- 
lars 1 

£       £  $ 

Statement, .  .  3  :  100  ::   10  :  $333,33^-  Ans. 

8.  In  £100,  New  Jersey  currency,  how  many  dollars  ? 

Ans.  $266,66|. 

9.  Reduce  54  pounds  6  shillings,  New  England  cur- 
rency, to  dollars  ?  A?is.  $181. 

10.  Reduce  304  pounds  9  pence,  New  England  cur- 
rency, to  dollars  ?  Ans.  $1013,458. 

11.  Reduce  364  pounds  2  shillings  6  pence,  New  York 
currency,  to  dollars  ?  Ans.  $910,31  + 


REDUCTION    OF    CURRENCIES.  191 


(Art.  109.)  The  reverse  operations  will,  of  course, 
bring  dollars  to  pounds.  In  New  England  currency, 
1  dollar  equals  f^^  of  a  pound;  in  New  York  currency, 
1  dollar  equals  y\  of  a  pound  ;  in  New  Jersey  currency, 
1  dollar  equals  f  of  a  pound  ;  hence,  to  reduce  dollars  to 
pounds,  we  must  take  the  following 

Rule.  Multiply  the  dollars  ayid  decimals  of  a  dollar 
by  the  value  of  a  dollar^  expressed  in  parts  of  a  pound. 

EXAMPLES. 

1.  In  45  dollars  75  cents,  how  many  pounds,  shillings 
and  pence,  New  England  currency  ? 
45,75 
,3 


13,725 
20 

14,500 
12 


6,000       Ans.  ^13  14  s.  Q  d. 
2.  Reduce  90  dollars  to  pounds,  <fec.  Georgia  currency. 
Operation. 
90 


90X/o'  or  30 


Ans.  ^621. 


3.  Reduce  640  dollars  to  pounds.  New  Jersey  cur- 
rency.    Thus, 


8 


640 
3  Jins.  £2i. 


4.  Reduce  500  dollars,  to  New  England,  New  York, 
New  Jersey  and  Georgia  currencies. 

f$500=£l50  New  England  currency. 
J  $500=^2200  New  York  currency. 
•^^^*  ]  $500=-£187   10  5.  New  Jersey  currency. 
L$500=^eil6  13  5.  4  f/.  Georgia  currency. 

[Note.  From  these  answers,  we  perceive  that  ^6150, 
New  England  currency,  is  equal  to   ^6200,  New   York 


192 


ARITHMETIC. 


currency;  and  in  the  same  manner,  we  may  compare  any 
one  with  any  other.  Also,  we  observe,  that  to  change 
New  England  currency  to  its  equivalent  value,  N.  York 
currency,  we  must  add  its  i ;  and  to  reduce  New  York 
to  its  value  expressed  in  New  England  currency,  we 
must  subtract  its  \.  But  such  problems  never  occur  in 
business.] 


REDUCTION  OF  FOREIGN  CURRENCIES  AND 
EXCHANGE. 

In  different  countries,  different  modes  of  reckoning  mo- 
ney always  have  existed,  and  no  doubt  always  will  exist; 
and  it  is  important  for  commercial  men  to  he  ready  and 
familiar  with  the  most  common  and  practical  operations  ; 
but  there  can  be  no  difficulty  with  those  who  understand 
the  general  principles  of  reduction ;  and  all  the  exam- 
ples of  money  exchanges  that  can  now  be  brought,  are 
but  illustrations  and  applications  of  those  principles  al- 
ready supposed  to  be  learned:  hence  we  shall  only  add 
a  few  examples  after  the  following 

TABLE  OF  FOREIGN  CURRENCIES. 


English  Money, 
^1=$4,44,    before    1832; 
since  1832, — 

£1=$4,80,  or,^5  =  $24,00 
1  English  crown  •  =$  1,10 
Or,  10  crowns  •  .  =-$11,00 
1  English  shilling  .  .  =,222 
Irish  Currency. 

£\  Irish, $4,10 

1  shilling, ,205 

1  penny, ,017 

French  Currency. 
French    crowns    same    as 
English. 


1  Five  franc  piece,  •  •  $0,93 

1  Franc, ,186 

1  Decime, ,0180 

Currencies  of  other  nations. 
The  Roman  crown,  Span- 
ish and  Mexican  dollar,  and 
the  rix-dollar,  of  Sweden, 
are  the  same  in  value  as  the 
dollar,  federal  money. 
1  Milree,  of  Portugal,  $1,24 
1  Russian  Rix-dollar,  ,600 
1  Ducat  of  Naples, .  .  .  ,80 
1  Florin,  of  Trieste,  •  •  ,48 
1  Rix-dollar,  of  Trieste,  ,96 
1  Rupee,  of  Bengal,  •  •  ,55^ 


REDUCTION  OF  FOREIGN  CURRENCIES.      193 


1  Rupee,  of  Bombay  •  •  ,50 
1  Tale,  of  Canton,  •  -1,48 
1  Tale,  of  Sumatra, .  •  4,1G 

Canada  Currency. 
1  shilling, ,20 


1  Florin,  of  Java,    ...  ,40 
1  Mark  Banco,  of  Ham- 
burg,   ,331- 

1  Guilder,   of    Amster- 
dam,   ,40 


* 


EXAMPLES. 

1.  Reduce  ^5  pounds  7  shillings  10  pence,  sterling  mo- 
ney, to  dollars  and  cents.  Solution  :  as  j6  1  =  480  cents,  1 
shiiling=24  cents,  and  1  penny=2  cents.  Therefore, 
45X480+7X24-1-10X2=21788  cents,  or,  $217,88. 

2.  Reduce  13  pounds  10  shillings  4  penc^e,  sterling  mo- 
ney, to  dollars,  cents,  &c.  £ns.  $64,88. 

3.  Reduce  13  pounds  13  shillings  G  pence,  sterling,  as 
valued  before  1832,  to  dollars  and  cents. 

Solution:   13,675 X  V  =^60,77 +,  Ans. 

(Art.  111.)  From  these  operations,  v/e  perceive  that, 
to  reduce  dollars  and  cents  to  pounds,  shillings  and  pence, 
sterling,  at  the  old  valuation,  we  must  first  inulliply  the 
dollars  and  decimal  by  ^''^ ;  and  secondly,  reduce  ike  de- 
cunal  part  to  shillings  and  pence.  For  the  modern  val- 
uation, we  have  the  ibllowing 

Rule.  Divide  the  sum  in  cents  by  2,  and  the  quotient 
is  pence,  ivhich  reduce  to  shillings  and  pounds.  In  all 
opercdiunSs,  cancel  ivhen  practicable. 

4.  [\\  480  dollars,  how  many  pounds  sterling,  as  valued 
prior  to  1832? 

Solution:  480X /„  =12X  9=£108,  Ans. 

5.  In  $720,46,  how  many  pounds,  shillings  and  pence, 
as  valued  since  1832? 

Operation. 
2)72046 


12)36023=peiice. 

20)3001  =  sliillings,  and  11  over, 

i^lSO  Is.  lid.  Ans. 
- 


I   194  ARITil.-IETIC. 


6.  Reduce  1200  dollars  to  pounds,  &c.,  sterling  money, 
as  now  valued.  c'i/i.S".  £250. 

7  Reduce  600  pound  10  shillings  sterling,  to  dollars 
and  cents.  ^^ns.  $2882,40. 

8.  In  47  pounds  2  shillings  6  pence,  Irish  currency, 
how.many  dollars  and  cents  ?  ^ns.  S193,21  + 

9.  Reduce  720  dollars  40  cents  to  pounds,  &c.,  Irish 
currency. 

410)72040(17512.^8;  or,  .6175  9  s.  3  d. -\~  ^ 

10.  In  3405  milrees  of  Portugal,  how  many  dollars  ? 

.^ns.  84222,20. 

11.  Reduce  6246  milrees  to  dollars. 

,.^ns.  $7745,04. 

12.  Reduce  248  dollars  50  cents  to  milrees  of  Portu- 
gal, ^^ns.  200^. 

13.  In  330  Russian  rix-dollars,  how  many  Spanish. 
Mexican,  or  Federal  dollars?  ^■9ns.  8220. 

14.  Reduce  960  dollars  to  Russian  rix-dollars. 

j3ns,  1440. 

15.  Reduce  5272  francs  to  federal  dollars. 

Jns.  S980,59  + 

16.  Reduce  1200  dollars  to  francs. 

.^ns.  6451,61  -f 

17.  In  2400  rupees  of  Bengal,  liow  many  dollars  ? 

.^ns,  $1335,33, 

18.  In  4282  ducats  of  Naples,  how  many  dollars  ? 

Ans,  $3425,60. 

19.  Reduce  5240  dollars  to  ducats  of  Naples. 

.^ns.  6550. 

20.  Reduce  320  dollars  to  marks  hanco  of  Hamburg. 

.'2?7.s.  960.  . 

21.  In  4560  tales  of  Canton,  how  many  dollars? 

.diis.  86748,80. 

22.  In  735  florins  of  Java,  or  guilders  of  Amsterdam, 
how  many  dollars  and  cents  ?  »^??s.  $294,00. 

23.  In  40  pounds  12  shillings  8  pence,  Canada  cur- 
rency, how  many  dollars  ?  .■^n.s.  $162,53;?-. 

24.  Reduce  740  dollars  42  cents,  to  pounds,  shillings 
and  pence,  Canada  currency,     .^ns.  ^6185  2  5.  lyV  d. 

25.  In  428  tales  of  Sumatra,  how  many  dollars  ? 

^ns.  $1780,48. 


EXCHANGE    AND    PER-CENTAGE.  195 


EXCHANGE  AND  PERCENTAGE, 

COMBINED. 

(Art.  112.)  By  some  writers,  tliis  is  simply  called  ex- 
change ;  but  we  hope  lo  be  more  clear  by  being  (in  this 
case)  moie  technical.  The  preceding  problems  come  lit- 
erally under  the  head  of  reduction  of  currencies,  which 
may  be  called  exchange  without  premium.  But  as  a  ge- 
neral thing,  when  it  is  necessary  to  reduce  money  from 
one  currency  to  another,  it  is  done  in  consequence  of  some 
commercial  transaction  which  requires  profit,  loss  or  com- 
mission, at  the  same  time.  Also,  in  the  same  operation, 
we  may  make  allowances  for  waste,  shrinkage,  reduc- 
tions for  weight  of  boxes,  bales,  &c.,  all  by  multiplica- 
tion and  division.  As  all  reductions  of  every  kind  and 
name  must  have  some  given  proportion,  and  reductions 
by  proportion  are  effected  by  multiplications  and  divis- 
ioi^.s  ;  therefore,  we  can  combine  all  these  difTerent  al- 
lowances under  one  general  operation,  as  in  Art.  24.  The 
allowances  for  weight  of  bales,  boxes,  &c.,  have  been 
called  tare  and  tret ;  but  this  technicahty  is  of  no  conse- 
quence. 

Under  this  head  the  problems  and  conditions  can  be  so 
varied,  tliat  definite  rules  would  only  be  cumbersome  : 
the  pupil  must  rely  on  general  principles  and  good  judg- 
ment.     "\Ye  shall  explain  by  the  following 

EXAMPLES. 

1.  Purchased,  in  London,  300  yards  of  broadcloth, 
which  cost  me,  including  transportation  and  duties,  260 
pounds.  I  must  meet  another  charge  of  2  per  cent,  for 
commission  ;  and  before  my  capital  returns  I  expect  the 
average  time  must  be  9  months,  and  money  is  worth  4 
per  cent. :  taking  these  circumstances  into  consideration, 
how  must  I  sell  the  cloth  per  yard,  in  Federal  Bluney, 
to  gain  1 5  per  cent.?  Ans.  $5,286  -f 

The  first  operation  is  to  reduce  the  260  pounds  to  dol- 
lars.    Draw  a  perpendicular  line,  (Art.  24,)  thus  ; 
260 
24 


196  ARiTHMrnc. 


Tliis  may  now  be  considered  as  done  ;  and  the  next 
step  is  to  increase  it  2  per  cent.,  and  that  is  done  by  mul- 
tiplying by  102,  and  dividing  by  100.  Then  we  shall 
have,  as  here  represented. 

260 


5 
100 


24 
102 


"We  must  now  give  interest  on  this  money,  at  4  per 
cent,  for  9  months  ;  that  is,  multiply  by  103,  and  divide 
by  100.  After  that,  we  must  increase  it  15  per  cent., 
and  we  shall  have  the  whole  amount  of  dollars  to  be  re- 
ceived for  the  300  yards  of  cloth.  We  then  divide  by 
300,  and  we  shall  have  the  sum  corresponding  to  one 
yard,  and  the  work  is  done,  all  but  the  canceling.  When 
completed,  in  form  it  stands  thus  : 

260 


5 

100 
100 
100 
300 


24 

102 
103 
115 


This  will  cancel  in  part,  and  as  far  as  it  will  cancel,  it 
is  so  much  labor  saved  ;  but  the  chief  merit  insisted  up- 
on, is  the  plan,  going  through  the  entire  operation  in  form 
without  multiplying  or  dividing,  going  through  the  solu- 
tion as  a  reasoner,  and  not  as  a  mechanical  ope'rator 
only. 

Let  this  be  a  jyattern  or  a  ride  for  all  other  problems 
of  the  like  kind. 

2.  A  merchant  bought  280  yards  of  sheeting  in  Liver- 
pool for  25  pounds,  and  duties  and  transportation  cost  7 
pounds  10  shillings  more  ;  what  shall  be  the  price  per 
yard  in  cents,  to  gain  20  per  cent,  on  first  cost  ? 

.4'??.?.  66  cents  8  mills. 

3.  Received  600  yards  of  Irish  linen,  the  whole  cost, 
including  transportation,  duties  and  commission,  was  75 
pounds  10  shillings,  Irish  currency  ;  what  must  be  my 
retail  price  per  yard,  in  Federal  Money,  to  gain  20  per 
cent.?  »^ns.  60,4  cents,  nearly. 


EXCHANGE    AND    PER-CENTAGE.  197 


4.  A  merchant  bought  sugar  in  New  York  at  6  pence 
per  pound,  New  York  currency  ;  and  while  on  his  iiand 
the  wastage  was  estimated  at  5  per  cent.,  and  interest  on 
tlie  first  cost  to  the  time  of  sale,  at  2  per  cent,  ;  how  ma- 
ny cents  shall  he  ask  per  pound  to  gain  25  per  cent.? 

Ans.  8^  cents. 

5.  A  merchant  bought  3  hogsheads  of  Lisbon  wine  at 
I  of  a  milree  per  gallon :  on  arrival  at  New  York,  it  was 
found  that  one  hogshead  had  lost  21  gallons  by  leakage  ; 
the  duties  and  transportation  amounted  to  18  per  cent,  on 
the  original  cost ;  what  shall  he  sell  it  per  gallon,  to  gain 
20  per  cent.?  Jins.  79  cents. 

6.  Received  from  Malaga  six  hogsheads  of  wine, 
which,  including  duties  and  transportation,  cost  me  175 
Spanish  dollars  :  it  has  suffered  a  leakage  of  about  4  per 
cent. ;  how  must  I  sell  it  per  gallon  to  gain  12  V  per  cent.? 

A71S.  54}  cents. 

7.  I  bought  560  yards  of  Irish  linen,  which  cost  me, 
including  all  expenses,  90  pounds  Irish  currency,  and  I 
sold  it  at  75  cents  per  yard  ;  what  was  my  gain  per  cent.? 

Ans.  14  per  cent,  nearly. 
cost.  sale. 

Solution:  as  90X410  :  560X75  ::  100  :  4th  term. 
8.  If  108  dollars  were  given  for  9  hundred  weight  of 
sugar,  nett  weight,  tare  16  pounds  per  hundred  weight, 
what  must  be  the  price  per  pound,  to  gain  12  per  cent.? 

Ans.  14  cents. 

9.  Received  from  my  agent,  300  barrels  of  flour,  which 
cost  me  1400  dollars,  and  75  dollars  incidental  expenses; 
I  now  ship  the  same  to  Brazil,  and  pay  $1,25  per  barrel 
for  transportation;  what  must  it  bring  in  that  market  to 
yield  me  15  per  cent.,  exclusive  of  duties  at  that  port? 

Ans.  S7,09  -I- 

10.  A  merchant  bought,  in  the  British  West  Indies,  3 
casks  of  molasses,  each  containing  54  gallons,  for  12  pounds 
13  shillings  6  pence;  transportation  and  duties,  ^63  4  s. 
6  d.  in  addition ;  he  then  estimates  a  wastage  of  4  per 
cent.;  what  shall  be  his  price  per  gallon,  in  federal  money, 
to  gain  20  per  cent.?  Ans.  ,58  +  cts. 

11.  What  is  the  cost  of  15  chests  of  tea,  each  contain- 

^  __ 


198  ARITHMETIC. 


ing  147  pounds  gross,  at  4  shillings  6  pence,  New  York 
currency,  per  pound,  tare  16  pounds  on  every  112  pounds? 

Ans.  $1063,12^. 

12.  Sold  cloth  at  12  shillings  6  pence,  New  York  cur- 
rency, and  gained  15  per  cent.;  what  would  have  been 
the  price  per  yard,  in  dollars  and  cents,  had  I  gained  27 
per  cent.?        "  .^ns.  $1,725  -f 

12,5X100 

Solution: ■  :     fl     ::     100    :     127 

8X115  "--^ 

Dividing  by  8,  reduces  shillings  to  dollars,  New  York  cur- 
rency. 

13.  A  merchant  received  from  Amsterdam,  620  yards 
of  carpeting,  whole  cost  2480  guilders ;  what  must  be 
the  retail  price  per  yard,  in  federal  money,  to  gain  15  per 
cent.?  Anb'.  $1,84. 

14.  jYew  York,  March  30,  1842.  This  day  received 
from  Liverpool,  900  yards  of  broadcloth,  whole  cost  1200 
pounds  sterling;  how  must  I  sell  it  per  yard,  in  federal 
money,  to  gain  20  per  cent.?  '  Aiis.-  $7,68. 

15.  Boston,  July  1,  1843.  This  day  received  from 
France,  1200  bottles  of  Champagne,  for  which  I  paid  400 
French  guineas,  each  $4,60;  how  must  I  sell  this  wine 
per  botde,  to  gain  35  per  cent.?  Ans.  $2,07. 

16.  "  Received  300  ells  of  cloth  from  Hamburg,  which 
cost  1500  marks  banco ;  how  must  the  same  be  sold  per 
yard,  in  federal  money,  to  gain  12 1  per  cent.,  the  ell 
Hamburg  being  2\  qr."  "         Ans.  $3,00. 

17.  A  man  shipped  flour  to  Dublin:  whole  cost,  in- 
cluding purchase,  transportation,  commission,  and  duties, 
was  4500  dollars  ;  he  directs  his  agent  to  sell  it  not  un- 
der an  advance  of  10  per  cent.;  how  many  pounds,  Irish 
currency,  could  he  draw  for?     Ans.  £1207  6  s.  4  d.-{- 

(Art.  113.)  Connected  with  the  higher  order  of  ex- 
changes, is  a  certain  class  of  problems  which  trace  a 
result  through  several  proportions:  yet  different  from 
compound  proportion,  as  that  which  is  generally  consid- 
ered compound  proportion,  is,  in  reality,  in  respect  to 
things,  single  proportion,  and  compound  only,  in  respect 
to  nwnericcd  factors  ;  but  the  proportions  we  are  about 


EXCHANGE    AND    PER-CENTAGE.  199 


to  introduce,  change  from  one  thing  to  another,  in  fact, 
and,  after  being  condensed  and  properly  linked  together, 
the  single  operation  is  called  Conjoined  Proportion :  or, 
The  Chain  Rule.  We  shall  treat  it,  however,  as  simple 
reduction ;  but  reduction  is  proportion,  as  we  may  see 
by  the  following  example  : 

Reduce  3  tons  to  pounds,  by  proportion ;  thus, 


First.     As 


ton. 

tons. 

cwt. 

1 

3     :: 

20 

20 

60 


Second 

As 

cwt. 

1 

cwt. 
:    60     :: 

4 

240  qr. 

qr. 
4    : 

Third. 

As 

qr. 
1     : 

qr. 
240    :  : 

28 

1920 

48 

6720  Ib.-- 

lb. 

28 

^Ans. 

Thus  we  perceive  that  reduction  is  proportion ;  but  the 
formal  statements,  one  by  one,  step  by  step,  render  all 
such  simple  problems  complex.  It  is  obvious  that  these 
three  proportions  are  linked  together  ;  they  are  conjoined, 
or  enchained ;  but,  as  a  whole,  are  entirely  unlike  com- 
pound proportion.  We  call  particular  attention  to  this, 
as  some  authors  have  m?de  the  sad  mistake  of  placing 
problems  in  conjoined,  under  rules  of  compound,  pro- 
portion. 

Let  us  return  to  our  example  again.  We  observe  that 
the  first  terms  of  the  several  proportions  are  units  :  hence 
there  will  be  no  dividing;  the  whole  operation  will  be 


200  ARITHMETIC. 


actual  multiplication,  and,  by  Art.  24,  the  entire  operation 
will  stand  thus  : 

3 
1     20 
1       4 
1     28 
This  may  be  read  as  follows:     *What  number  of 
pounds  in  3  tons,  if  1   ton  =20  cwt.,  1  cwt.=4  quar- 
ters, and  1  quarter=28  pounds  ?     We  may  render  this 
same  question  much  more  complex,  M-ithout  changing  its 
nature.     It  will  still  be  reduction,  stated  thus  : 

If  2  tons  in  London  make  40  hundred  weight  in  New 
York,  and  3  hundred  weight  in  New  York  make  12  quar- 
ters in  Havana,  and  4  quarters  in  Havana  are  equal  to  112 
pounds  in  Brazil,  how  many  pounds  in  Brazil  are  equal 
to  3  tons  in  London  ? 

This  solution  will  stand  thus  : 

3  tons,  term  of  demand. 
If  tons,   .    2=  40  hundred  weight; 
and  cwt.,    3=    12  quarters; 
and  qr.,   .  4=112  pounds. 
Divide  or  cancel  out  the  numbers  on  the  left  hand  side  of 
the  line,  and  it  will  be  the  same  as  before  expressed. 

This  is  rendered  complex,  by  mentioning  places, 
which  need  not,  yea,  should  not  be  mentioned  in  connec- 
tion with  this  problem,  but  which  must  be  mentioned  in 
connection  with  some  others  ;  we  mention  them  here,  to 
show  how  mere  names  of  places,  or  different  names  to 
things,  will  throw  confusion  into  the  mind,  when  in  real- 
ity the  problem  may  be  simple. 

For  a  further  illustration,  let  us  take  the  following  well 
known  problem. 


*  In  view  of  these  equalities,  and  the  subsequent  arrangements  of 
the  terms  each  side  of  a  perpendicular  line,  some  have  been  led  to 
say  that  tliis  perpendicular  line,  in  all  cases,  represents  equality — an 
unphilosophical  assertion.  Indeed,  it  is  to  be  apprehended  that  this 
whole  system  of  canceling  may,  for  a  time,  suffer  in  the  estimation 
of  the  public,  by  the  immodest  pretensions  of  some  of  its  pro- 
fessors.   


EXCHANGE    AND    PER-CENTAGE.  201 


If  50  pounds  Englisli,  make  45  pounds  Plcmish,  and 
22  pounds  Flemish,  make  28  pounds  Bologna,  how  ma- 
ny pounds  English,  are  equal  to  5G  pounds  Bologna? 

Ans.  48 1. 
This  may  be  done  by  two  statements  in  proportion, 
thus : 

FL     Fl.     F)ig.     Eng. 
22  X  50 

As  45  :  22  : :  50  : =28  lb.  Bologna. 

45  ^ 

22X50 

Again,  as  28  :  56  : : :  Answer. 

45 


56  term  of  demand,  Bologna. 

22  Flemish. 

50  EnMish,  odd  term. 


ByArt.24.^P^^^_^ 

By  paying  attention  to  the  order  of  arrangement  of  the 
terms  in  the  solutions  of  the  preceding  problems,  the  pu- 
pil will  understand  the  rationale  of  the  following 

Rule.  Distinguish  the  terms  or  condition /rom  the 
one  o/' DEMAND.  Draw  a  perpendicidar  line  ;  first,  and 
to  the  right  of  it  place,  the  term  of  demand.  Place  the 
conditional  term  of  the  same  name  in  a  line  below  and 
to  the  left  of  the  perpendicidar,  and  opposite  to  it  on  the 
right  put  the  term  ecjucd  to  it.  On  the  left  agcmi,  place 
the  conditional  term  of  the  same  name,  as  the  last  one 
on  the  right;  and  so  on,  until  all  the  terms  are  nsed. 
The  last  term  on  the  right  will  be  cm  odd  term,  of  the 
same  name  as  the  answer. 

Then  cancel  down,  and  the  resxdt  of  the  division 
of  the  right  hand  factors  by  those  on  the  left  hand 
of  the  line,  will  be  the  answer  sought. 

N.  B.  If  we  had  the  answer  at  first  to  put  on  the  left 
of  the  perpendicular  line,  the  product  of  the  factors  on 
both  sides  would  then  balance. 

EXAMPLES. 

1.  If  12  pounds  at  Boston  make  10  at  Amsterdam,  and 
100  at  Amsterdam,  are  equal  to  120  pounds  at  Paris; 


202  ARITHMETIC. 


Solution : 

B.  40 
H.  90 


liow  many  pounds  nt  Boston  are  equal  to  100  pounds  at 
Pai-is  ?  -'^'f'"'  1^0  pounds. 

Solution—  100  pounds  at  Paris,  term  of  demand  ; 

Paris,.  .  •  120      100  pounds,  Amsterdam  ; 
Amsterdam,   10        12  pounds,  Boston,  odd  term. 

2.  If  14  braces  at  Venice  be  equal  to  15  braces  at  Leg- 
horn, and  7  braces  at  Leghorn  equal  to  4  American  yards, 
how  many  braces  of  Venice  are  equal  to  30  American 
yards  ?     "  ^ns.  49. 

3.  If  40  pounds  at  Boston  make  36  in  Holland,  and 
90  pounds  in  Holland  make  116  in  Poland,  how  many 
pounds  in  Poland  are  equal  to  250  at  Boston? 

£ns.  290  lbs. 
250  demand,  B. 
36  H. 

116  P.  (odd  term.) 
The  pupil  will  perceive  tliat  the  odd  term,  in  every  case, 
is  balanced  by  the  answer,  when  found. 

4.  A  merchant  in  Russia  owes  950  ducats  in  Berlin, 
which  he  wishes  to  pay  by  an  order  for  rubles,  by  the 
way  of  Holland  ;  he  has  this  information,  that  2  rubles 
make  95  stivers,  and  20  stivers  make  1  florin,  and  5  flo- 
rins make  2  rix-dollars  of  Holland;  100  rix-doilars  of 
Holland  are  equal  to  142  rix-dollars  of  Prussia,  and  3 
rix-dollars  of  Prussia  are  equal  to  1  ducat  of  Berlin ;  how 
many  rubles  will  pay  the  debt?     .^??s.  21124f  rubles. 

5.  If  A  can  do  as  much  work  in  3  days  as  B  can  do 
in  ih  days,  and  B  can  do  as  much  in  9  days  as  C  in  12, 
and  C  as  much  in  10  days  as  D  in  8,  how  many  days' 
work  of  D  are  equal  to  5  days  of  A  ?  .4^5.  8. 

6.  If  2  pounds  of  tea  are  worth  11  pounds  of  coflee, 
and  3  pounds  of  coffee  worth  5  pounds  of  sugar,  and  18 
pounds  of  sugar  worth  21  pounds  of  rice,  how  many 
pounds  of  rice  can  be  purchased  with  12  pounds  of  tea? 

./???5.  1281. 
We  found  the  following  under  the  rule  of  compound 
proportion;  but  we  think  it  wandered  out  of  its  place: 

7.  "  If  25  pears  can  be  bought  for  10  lemons,  and  28 
lemons  for  8  pomegranates,  and  1  pomegranate  for  48  al- 
monds, and  50  almonds  for  70  chestnuts,  and  108  chest- 


FKLLOWSHIP.  203 


nuts  for  2\  cents,  how  many  pears  can  I  buy  for  135 
cents?"      "  . 'i 22V.  759a.  ^ 

8.  If  7  pounds  of  raisins  are  worth  10  pounds  of  cof- 
fee, and  10  pounds  of  coffee  worth  3  g-nllons  of  molasses, 
and  2  gallons  of  molasses  worth  62 .'^  cents,  how  many 
pounds  of  raisins  can  be  bought  for  8  dollars? 

,/lns.    59\l  lbs.   nearly. 


FELLOWSHIP. 


(Art.  114.)  Fellowships  are  of  two  kinds,  single  and 
double;  the  former  includes  such  cases  as  take  no  account 
of  time;  the  latter  admits  time  as  an  element  in  the  trans- 
action. When  men  associate  for  any  single  act  of  spec- 
ulation or  adventure,  or  when  they,  in  con>mon,  are  to 
sustain  a  specific  loss,  or  raise  a  specific  tax,  tlie  opera- 
tion of  determining  each  man's  portion  of  the  gain  or 
loss,  is  called  single  fellowship.  This  is  not  applicable 
to  business  partnerships,  as  generally  conducfcd,  in 
which  a  fixed  proportion  of  the  profit  or  loss  is  usually 
agreed  upon  between  the  parties,  and  the  disparity  in  the 
amount  of  capital  is  compensated  by  personal  services, 
or  by  an  allowance  of  interest  on  the  excess  furnished  by 
either.  In  single  transactions,  such  as  fall  under  single 
fellowship,  it  is  plain  that  he  who  furnished  most  capital, 
either  iu  money  or  labor,  must  have  most  of  the  profit; 
and  he  who  hazards  most  in  any  adventure,  must  sustain 
most  of  the  loss,  when  loss  occurs  ;  and  just  in  propor- 
tion to  that  capital,  or  tliat  hazard,  compared  to  the 
capital  or  hazard  of  others.  Hence  fellowship  is  but 
another  application  of  that  pure  2>rinciple  of  pro- 
portion, which,  as  applied  to  such  cases,  may  be  express- 
ed thus : 

As  the  whole  amount  of  stock  or  labor 

Is  to  each  man's  portion. 

So  is  the  whole  property,  loss  or  gain, 

To  each  man's  share  of  it. 
Proof.    The   sum   of  all   the  shares  must  equal  the 
whole  gain,  &c. 


204 


ARITHMETIC. 


exa:«ples. 
^l.  Three  workmen  having  undertaken  to  do  a  piece  of 
work  for  275  dollars,  agreed  to  divide  their  protits  in 
proportion  to  the  amount  of  labor  each  one  performed. 
M  labored  50  days,  N  65  days,  and  O  85  days  ;  what 
was  the  share  of  each? 
Solution:  •  •  M  =  50  days. 
N=:65     " 
0=85     " 


Sum, 200 

Divide  by  25,  and      8 

Also, 8 

And 8 


:  $275 

:  50  : 

:        11 

:  50  : 

:        11 

:  65  : 

:        11 

:  85  : 

$68,75  =M's 

89,37?,  =  N's 

116,87i  =  0's 


$275,00  Proof. 
2.  A  man  bequeathed  to  the  eldest  of  his  four  children 
2400  dollars  ;  to  the  second,  2000  dollars  :  to  the  third, 
1800  dollars,  and  to  the  youngest,  1600  dollars,  or  in  that 
proportion  of  the  value  of  his  real  estate;  but,  on  the  set- 
tlement, only  4000  dollars  were  found  to  divide ;  what 
was  the  share  of  each  ? 

N.  B.  It  is  not  necessary  to  use  the  identical  numbers 
given  ;  it  is  sufficient  to  use  numbers  that  bear  the  same 
relation.     Hence  we  take, 
1st.    12 
2d.     10 
3d.      9 
4th.     8 


39  :  4000 


12 


123J3-  dollars  =  1st. 


*  Here  we  compare  days'  work  with  dollars.  It  has  been  asserted 
by  many,  in  modern  times,  that  we  cannot  compare  unhke  things, 
and  that  such  comparisons  are  absurd,  &c.;  but  we  think  these  views 
over  nice,  at  least.  It  is  true  we  cannot  measure  unlike  things  by 
each  other,  but  we  can  compare  their  numerical  vahies.  In  the  pre- 
sent instance,  who  can  doubt  the  philosophy  of  comparing  the  days' 
work  to  the  dollars  received,  to  see  whether  one  is  more  or  less  innu- 
vierical  amount,  than  the  other]  Yet,  as  a  general  thing, we  admit 
it  is  more  clear  to  compare  quantities  of  like  kind;  and  while  we  make 
this  admission,  we  do  not  concede  that  it  is  more  philosophical. 


FELLOWSHIP.  205 


3.  A  man  is  indebted  to  A  $250,50,  to  B  $500,  to 
C  $349,50;  but  when  he  comes  to  nmke  a  settlement,  it 
is  found  he  is  worth  but  $960;  how  much  will  each  one 
receive,  if  it  be  in  proportion  to  their  respective  claims? 

fA  $218,618-1- 

Ans.   \  B  $436,363-f 

(^C  $305,018-1- 

4.  Four  men  hired  a  coach  to  convey  them  to  their  re- 
spective homes,  which  were  at  distances  from  the  place 
of  starting  as  follows:  A's  16  miles,  B's  24  miles,  C's 
28  miles,  and  D's  36  miles  ;  what  ought  each  to  pay,  as 
his  part  of  the  coach  hire,  which  was  13  dollars? 

Ans.  A  $2,  B  $3,  C  $3,50,  and  D  $4,50. 

5.  Divide  360  dollars  in  the  proportion  of  2,  3  and  4. 

Ans.  80,  120,  160. 

6.  In  the  year  1835,  three  men,  P,  Q,  and  R,  made  a 
joint  capital  of  7500  dollars,  to  purchase  city  lots  in  Buf- 
falo; P  put  in  1500  dollars,  Q  2500,  and  R  the  remain- 
der ;  ihey  soon  gained  paper  obligations  to  the  amount  of 
18000  dollars;  what  was  each  partner's  share  of  the 
gain?  rP=$3600. 


rp=i 

LR=! 


Ans.  <  Q=$8000. 

:$8400. 

(Art.  115.)  Taxes.  The  taxable  inhabitants  of  any 
city,  town,  or  county,  are  all  in  political  fellowship,  for 
the  payment  of  taxes  and  support  of  government.  A 
small  tax  in  some  states,  and  in  most  foreign  governments, 
is  imposed  on  every  male  citizen  above  a  certain  age, 
and  being  the  same  to  all,  independent  of  property,  is 
called  per  head,  or  per  poll,  or  poll  tax.  In  addition  to 
this,  tliere  is  a  tax  on  property,  both  fixed  and  movable  : 
i.  e.  real  and  personal :  the  amount  of  which  property  is 
determined  by  official  appraisers,  called  assessors,  and 
their  list  of  persons,  and  property  held  by  each,  is  called 
the  assessment-roll.  All  persons  on  this  roll  are,  as  we 
before  observed,  in  felloivship.  When  the  amount  of 
tax  is  determined  upon,  (the  poll-tax  of  the  district,  if 
any)  is  taken  out,  and  the  remainder  divided  out  among 
the  inhabitants,  in  proportion  to  their  division  of  property. 


20G 


ARITHMETIC. 


EXAMPLES. 

I.  If  a  t:ix  of  3000  dollars  be  assessed  upon  all  the 
taxable  property  of  a  toM'ii,  the  valuation  of  which,  on 
the  assessment-roll,  is  600,000  dollars,  what  will  it  be  on 
1,  2,  3,  4,  &c.,  to  10  dollars  ?  what  will  it  be  on  100  and 
1000  dollars? 

Solution:  As  600000  :  3000  ::  1   :  4th  term. 


Or,  as  200  :  1 

::  1  :  _i-.-=,005  =  5  mills. 

r$l  pays 

,005 

$6  pays  ,03 

2   " 

,01 

7  "   ,035 

Ans.  < 

3   " 

,015 

8  "   ,04 

4   " 

,02 

9  "   ,045 

L  5   '* 

,025 

10  "   ,05 

100  "   ,50 

1000  "  $5,00 

The  property  of  John  Wells,  in  this  town,  is  assessed 
at  $2780  ;  what  is  his  property  tax  ? 

As  $1000  pays  $5,     $2000  pays  $10,00 

700  pays       3,50 

80  pays         ,40 


Therefore, 2780  pays  $13,90,  Ans, 

In  the  same  manner,  estimate  the  tax  of  any  other  indi- 
vidual inhabitant  who  possesses  taxable  property. 

2.  The  corporation  of  a  city  vote  to  raise  an  extra  tax 
of$5400  for  certain  public  improvements,  and  the  assess- 
ment-roll sums  up  to  $2,700,000  ;  what  is  the  extra  tax 
on  a  dollar,  and  liow  much  extra  tax  must  A  pay,  who  is 
taxed  for  $3757  ? 

Ans.  2  mills  on  a  dollar:  A  pays  $7,515. 

3.  A  tax  of  3000  dollars  is  to  be  raised  in  a  certain 
town,  where  the  taxable  property  amounts  to  120000  dol- 
lars, and  the  number  of  polls  320,  at  50  cents  each  ; 
what  is  paid  on  a  dollar,  and  what  is  A's  tax  on  a  pro- 
perty valuation  of  2500  dollars  ? 

Ans.  ,023|  on  $1  ;  A's  tax,  $59,16|. 


COMPOUND    FELLOWSHIP.  207 


COMPOUND  FELLOWSHIP. 

(Art.  116.)  There  are  certain  cases  of  fellowship  that 
compel  lis  to  take  time  into  consideration.  The  produc- 
tive vaUie  of  stock  depends  on  the  amount  of  capital,  by 
the  time,  as  in  interest,  or  the  amount  of  labor  depends 
on  the  number  of  laborers,  by  the  time  they  labor,  &c. 
All  such  cases  must  be  considered  under  a  compound,  of 
which  time  is  an  element;  hence  it  is  called  compound 
fellowship.  The  following  example  will  fully  illustrate 
this  principle : 

Three  men  hired  a  pasture  lot  for  25  dollars  ;  A  put 
in  5  cows,  12  weeks;  B  4  cows,  10  weeks;  and  C  3 
cows,  15  weeks;  what  part  of  the  25  dollars  shall  ench 
pay  ? 

Here  it  is  evident,  that  the  number  of  cows,  each  put 
in,  cannot  decide  the  question,  nor  can  the  time  alone  de- 
cide ;  we  must  take  tJie  compound  of  both  these  elements, 
thus  : 

5  cows  12  weeks =60  cows  1  week. 
4  cows  10  weeks =40  cows  1  week. 
3  cows  15  weeks =45  cows  1  week. 

The  whole  pasturage  is,  then,  the  same  as  145  cows  1 
week ;  and  the  time  is  thus  reduced  to  unity,  and  then 
solved  by  single  fellowship;  thus, 

145  :  25  ::  60  :  A's  part. 

or, .  .  29  :  5  : :  60  :  A's  part.  A=$10,344  -{- 

29  :  5  ::  40  :  B's  part.  B=     6,897  + 

29  :  5  ::  45  :  C's  part.  C=     7,758  4" 

Froof,  $24,999  + 
lit  the  same  way,  we  estimate  labor,  as  well  as  con- 
sumption, interest,' &c.,  and  from  this  we  draw  the  fol- 
lowing 

Rule.  Multiply  the  active  ager.ts  by  the  time  each 
was  in  action;  or,  if  the  agent  is  money  or  stock,  mul- 
tiply each  man's  stock  by  its  time,  and  add  the  several 
products  together.     Then  by  proportion, 


208  ARITHMETIC. 


As  the  sum  of  the  products 

Is  to  each  particular  product, 

So  is  the  wJiole  gain  or  loss 

To  each  man's  share  of  the  gain  or  loss. 

EXAMPLES. 

1.  Three  merchants  traded  together;  A  put  in  120 
dollars  for  9  months,  B  100  dollars  for  16  months,  and  C 
100  dollars  for  14  months,  and  they  gained  100  dollars; 
what  is  each  man's  share  ? 

$      mo. 
A's  stock  120X    9  =  1080 
B's  stock  100X16=1600 
C's  stock  100X14  =  1400 


Sum,  4080 
Then,  as  4080  :   1080  ::   100 
or,  as  ...  102  :  27  : :   100  :  $26,47  -f  A's  share,  &c. 

2.  Two  men  commenced  partnership  for  a  year;  one 
put  in  1000  at  the  commencement,  the  other  put  in  his 
capital  4  months  afterwards  ;  at  the  close  of  the  year 
they  divided  equally  ;  what  capital  did  the  second  put  in  ? 

Ans.  $15tl0. 
N.  B.  Many  problems  must  be  solved,  rather  by  the 
general  principle,  than   by  the  general  rule.      In   the 
above  example,  if  they  shared  equally,  the  products  of 
their  capital  and  time  must  be  equa4. 

3.  A,  B,  and  C,  made  a  stock  for  12  months;  A  put 
in,  at  first,  873  dollars  60  cents,  and  4  months  after,  he 
put  in  96  dollars  more  ;  B  put  in,  at  first,  979  dollars  20 
cents,  and  at  the  end  of  7  months,  he  took  out  206  dol- 
lars 40  cents ;  C  put  in,  at  first,  355  dollars  20  cents,  and 
3  months  after,  he  put  in  206  dollars  40  cents,  and  5 
months  after  that,  he  put  in  240  dollars  more.  At  the 
end  of  12  months,  their  gain  is  found  to  be  3446  dollars 
40  cents  ;  what  is  each  man's  share  of  the  gain  ? 

rA's  share  is  $1362,88 
.^ns.  -i  B's  1298,34-j- 

[C's  785,13 

4.  Three  men  took  a  field  of  grain  to  harv'est  and 
thrash,  for  \  of  the  crop;  A  furnished  3  hands  4  days, 


COMPOUND    FELLOWSHIP.  209 


B  2  hands  5  days,  and  C  4  hands  6  days  ;  the  whole 
crop  amounted  to  320  bushels ;  what  was  each  partner's 
share  ? 

Ans, 


rA's=20|f. 
J  B's  =  17i'3- 
(.C's=4Ui. 


5.  A,  B,  and  C,  had  a  capital  stock  of  5762  dollars. 
A's  money  was  in  trade  5  months,  B's  7,  and  C's  9 ; 
they  gahied  780,  which  was  divided  in  the  proportion  of 
4,  5,  and  3 ;  that  is,  A  4,  as  often  as  B  5,  and  C  3 ;  now 
B  received  2087  dollars,  and  absconded ;  what  did  each 
gain,  and  put  in  ;  and  what  did  A  and  C  gain  and  lose  by 
B's  misconduct? 

f  A's  stock  $2494,887  gain  260 
^  J  B's  do.  2227,577  do.  325 
^^^'  1  C's  do.       1039,536         do.  195 

LA  and  C  gain $465,577 

[Note.  As  money,  in  compound  fellowship,  is  divided 
in  proportion  to  each  man's  capital,  multiplied  by  its  time 
in  trade;  inversely,  then,  each  man's  capital  must  be  in 
proportion  to  his  gain,  divided  by  the  time  his  money 
was  in  trade.  Therefore,  as  their  shares  of  the  gain  are 
as  the  numbers  4,  5,  and  3,  their  capital  stocks  must  be 
in  proportion  to  |-,  4  and  i.] 

6.  In  a  certain  factory  were  employed  men,  women  and 
boys ;  the  boys  receive  3  cents  per  hour,  the  women  4, 
and  the  men  6 ;  the  boys  work  8  hours  a  day,  the  wo- 
men 9,  and  the  men  12 ;  the  boys  receive  5  dollars  as 
often  as  the  women  10  dollars,  and  for  every  ten  dollars 
paid  to  the  women,  24  dollars  were  paid  to  the  men ;  how 
many  men,  women  and  boys  were  there,  the  whole  num- 
ber being  59  ?    Jlns.  24  men,  20  women,  and  15  boys. 

7.  A,  B,  and  C,  united  with  1930  dollars ;  A's  money 
was  in  trade  3  months,  B's  5,  and  C's  7 ;  they  gained 
117  dollars,  which  was  so  divided  that  |  of  A's  share 
was  equal  to  ^  of  B's,  and  to  \  of  C's ;  what  did  each 
gain,  and  put  in  ? 

^       C  A  gained  $26,  B  $39,  and  C  $52. 
*^"*'  I  A's  capital  was  $700,  B's  $630,  and  C's  $600. 

8.  A,  B,  and  C,  are  employed  to  do  a  piece  of  work 
for  26  dollars  45  cents ;  A  and  B  together  are  supposed 

72  """" 


210  ARITHMETIC. 


to  do  ^  of  it ;  A  and  C  j\,  and  B  and  C  i|,  and  are  paid 
proportionally  to  this  supposition ;  what  is  each  man's 
share  of  the  money  ? 

^7is.  A  $11,50,  B  $5,75,  C  $9,20. 


BARTER. 

(Art.  117.)  Barter  is  striking  the  balance  of  trade, 
or  determining  how  much  of  one  article  is  equal  to  a  giv- 
en quantity  of  another,  or  several  others;  and,  of  course, 
involves  no  new  principle  of  computation.  The  nature 
of  the  case  before  us  must  determine  the  mode  of  opera- 
tion.    Rules.     Practice.     Rule  of  Three. 

EXAMPLES. 

1.  A  farmer  bartered  3  barrels  of  flour,  at  5  dollars  25 
cents  per  barrel,  for  sugar  and  coflee,  to  receive  an  equal 
quantity  of  each;  how  much  of  each  must  he  receive, 
admitting  the  sugar  to  be  valued  at  9  cents  per  pound, 
and  the  coffee  at  15  cents  ?  ^^n.s.  65|. 

2.  Sold  75  barrels  of  herrings,  at  2  dollars  75  cents 
per  barrel,  for  which  I  am  to  receive  75  bushels  of  wheat, 
at  1  dollar  8  cents  per  bushel,  and  the  residue  in  money; 
how  much  money  must  I  receive?  ^ns.  $125,75. 

3.  Sold  35  yards  of  domestic,  at  20  cents  per  yard, 
and  am  to  receive  the  amount  in  apples,  at  25  cents  per 
bushel ;  how  many  bushels  must  I  have  ? 

^is.  28  bushels. 

4.  What  is  rice  per  pound,  when  340  pounds  are  given 
for  4  yards  of  cloth,  at  4  dollars  25  cents  per  yard  ? 

^ns.  5  cents. 

5.  Gave  in  barter  65  pounds  of  tea  for  156  gallons  of 
rum,  at  33^  cents  per  gallon ;  what  was  the  tea  rated  at  ? 

.^ns.  80  c.  per  lb. 

6.  How  many  pounds  of  butter,  at  12^  cents  per 
pound,  must  be  given  in  exchange  for  9|  pounds  of  In- 

I  digo,  at  235  cents  per  pound  ?  ^?is.  183y\  lb. 


BARTER.  21 


7.  A  merchMnt  bought  80  yards  of  broadcloth,  at  3-^^ 
dollars  per  yard,  and  paid  in  coifee  at  10  pence  per  pound, 
New  York  currency,  how  many  pounds  did  it  require? 

Arts.  2496  pounds, 

8.  Sold  24  hundred  weight  of  hops  at  4  pence  per 
pound,  and  took  pay  in  sugar  at  6  pence  per  pound  ;  how 
many  pounds  did  I  require?  Ayis.  16  cwt. 

9.  Sold  5  hundred  weight  1  quarter  of  sugar  at  8  dol- 
lars 50  cents  per  hundred  weight,  and  received  for  pay 
21  yards  of  cloth;  what  was  the  cloth  per  yard? 

Jlns.  $1,12^. 
All  the  examples  thus  far,  except  No.  2,  can  be  solved 
very  briefly  by  Art.  24.     No.  2  is  also  very  brief,  but 
not  in  that  form. 

10.  Q  has  coffee  worth  16  cents  per  pound,  but  in  bar- 
ter, raised  it  to  18  cents;  B  has  broadcloth  worth  4  dol- 
lars 64  cents  per  yard ;  what  must  B  raise  his  cloth  to, 
so  as  to  make  a  fair  barter  with  Q  ?  Ans.  $5,22. 

11.  P  and  Q  barter;  P  has  Irish  linen,  worth  3  shil- 
lings 7  pence,  cash,  but,  in  barter,  he  will  have  3  shillings 
10  pence  ;  Q  delivers  him  broadcloth  at  1  pound  16  shil- 
lings 6  pence  per  yard,  worth  only  1  pound  13  shillings; 
how  much  linen  does  P  give  Q  for  138  yards  of  broad- 
cloth, and  which  gains  by  the  bargain  ?  and  how  much 
in  dollars  and  cents,  the  whole  being  in  Irish  currency  ? 

Jins.  P  gives  1314  yards  ;  Q  gains  $  33-|-  »  because 
his  cloth  is  proportionally  too  high. 

12.  A  and  B  barter;  A  has  150  gallons  of  brandy,  at 
7  shillings  6  pence  per  gallon,  cash,  but  in  barter  he  will 
have  8  shillings ;  B  has  linen  at  3  shillings  6  pence  per 
yard,  ready  money;  how  much  must  it  be  per  yard,  to 
meet  A's  bartering  price,  and  how  many  yards  are  equal 
to  A's  brandy  ? 

Ans.  Barter  price,  3  s.  8f  d.;  B  must  give  A  321i 
yards,  nearly. 


212 


ARITHMETIC. 


SECTION  V. 


MENSURATION.* 

(Art.  118.)  Mensuration  is  finding,  or  defining,  the 
numerical  contents  of  surfaces  and  solids,  and  its  founda- 
tion is  the  science  of  geometry;  but  its  every-day  appli- 
cation to  the  mechanical  business  of  life,  demands  a  place 
for  its  practical  rules  in  every  arithmetic.  Some  of  these 
rules  are  obvious,  from  inspection:  otiiers  must  be  taken 
on  trust,  until  the  pupil  studies  geometry,  from  whence 
they  are  drawn  ;  we  shall,  however,  give  some  of  the 
common  illustrations.  All  surfaces,  of  whatever  figure, 
are  referred  to  squares  for  measurement,  and  the  iinit 
may  be  taken  at  pleasure ;  it  may  be  an  inch,  a  foot,  a 
yard,  a  rod,  a  mile,  &c.,  according  as  convenience  and 
common  sense  may  dictate.  The  following  figures  will 
render  the  truth  of  several  rules  apparent: 


Fig.  1 .  A  square. 


Fig.  2.  A  right  angled 

parallelogram. 
12      3    4    5    6     7 


The  first,  a  square,  the  side  of  which  is  3  units  in 
length ;  and  it  is  apparent  to  the  eye,  that  the  whole 
square  contains  9  square  units  ;  and,  considering  the  3 


*  Most  writers  on  arithmetic  have  placed  mensuration  after  the 
square  and  cube  roots ;  but  we  think  it  an  injudicious  arrangement, 
as,  in  the  operations  of  finding  roots,  mensuration  is  almost  univer- 
sally used. 


MENSURATION. 


213 


units  in  a  side  as  feet,  the  whole  square  is  one  square 
yard,  containing  9  square  feet,  agreeably  to  the  dictation 
of  the  square,  measure  table. 

By  figure  2nd,  we  perceive  that  the  number  of  litde 
squares  it  contains,  is  equal  to  the  iniits  in  lengtli,  mul- 
tiplied by  the  units  in  breadth,  which  must  be  true  in 
every  case. 

(Art.  119.)  If  the  sides  are  not  at  right  angles,  as  in 
figure  3d,  we  must  multiply  its  length  by  its  jierpendic- 
idar  breadth,  as  it  is  proved  in  geometry,  that  the  fiffure 
ABCG  =  L!ie  figure  BCDE. 

Fig.  3.  A  right  and  an  oblique-angled  parallelogram. 
A  E  G  D 


A  triangle  is  equal  to  half  of  a  parallelogram,  of  the 
same  base  and  altitude,  as  is  obvious,  from  inspecting  fig- 
ure 4,  AC=the  base,  and  BE=the  altitude. 

B  D 


Fig.  4. 


A  plain  figure  having  4  sides,  and  2  of  them  parallel 
and  unequal^  is  called  a  trapezoid.  It  is  rigidly  proved, 
in  geometry,  that  the  area  of  such  a  figure  is  found  by 
multiplying  its  perpendicular  altitude  EF,  by  a  medium 
base  between  AD  and  BC.  This  truth  is  apparent, 
from  the  inspection  of  Fig.  5. 

BEG 


D 


2U 


ARITHMETIC. 


Fig.  6. 

/                          g\ 

\                     ^ 

; 

(Art.  J20.)  When  a  circle  is  inscribed  in  a  squaie,  it 
lias  been  ascertained  by  geometry,  that  the  area  or  sur- 
face of  the  circle  occupies  ,7854  (as  near  as  can  be  ex- 
pressed with  4  decimals,)  of  the  whole  square.  Bu*  the 
square  is  the  square  of  the  diameter  of  the  circle,  '«s  ap- 
pears from  figure  6. 


Therefore,  to  find  the  area 
of  a  circle,  we  may  square 
the  diameter,  and  multiply  it 
by  the  decimal  ,7854. 


It  is  also  shown  in  geome- 
try, that  the  circumference 
of  a  circle  is  3,1416,  nearly, 
when  the  diameter  is  1 ;  and 
when  the  diameter  is  2,  the 
circumference  will  be  double, 
and  so  on  in  proportion :  ten  times  in  diameter  giving 
ten  times  for  circumference,  &c.  &lc. 


Again:  in  figure  6,  from  the  centre  C,  we  can  draw 
radii  CD,  CG,  as  near  each  other  as  we  please;  and 
if  we  draw  them  extremely  near  each  other,  the  small 
part  of  the  circumference  GD,  may  be  considered  a 
straight  line,  and  the  sector  CDG,  a  triangle  ;  but  the 
area  of  this  triangle  is  equal  to  CD,  multiplied  by  \  of 
GD.  But  the  whole  circle  can  be  filled  up  with  such 
triangles  :  hence  the  area  of  the  whole  circle  is  equal  to  V 
of  its  diameter,  multiplied  by  \  of  all  the  GD's,  or  ^  of 
the  whole  circumference.  Squares,  circles,  and  all  fig- 
ures of  a  similar  shape  to  one  another,  are,  in  area,  as 
the  square, of  their  like  dimensions.  That  is,  a  circle  of 
double  diameter  to  another,  is  4  times  in  surface,  of  .3  times 
in  diamefe^-r,  9  times  in  surface,  &c.  &:c.  This  may  be 
brought  to  the  mind  by  inspecting  figure  1.  If  we  call 
the  whole  square  in  fiffure  1  an  unit,  then  the  square  of 
1  is  J^';  that  is,  to  square  a  fraction,  square  its  numera- 
tor and  denominator:  geometrical  figures  and  numericals 
agree. 


.-VIE.NSl  KATIOX.  215 


RIXAPITULATION. 

To  find  the  surface  or  area  of  a  square. 

Rule  1.  Multiply  its  equal  length  and  breadth  to- 
gether. 

To  find  the  area  of  a  parallelogram. 

Rule  2.  Multiply  length  and  breadth  together,  (ta- 
king care  always  to  measure  them  at  right  angles.) 

To  find  the  area  of  a  triangle. 

Rule  3.  TVlien  the  base  and  altitude  are  given,  mul- 
tiply one  by  half  the  other. 

When  not  a  right  angle,  and  the  altitude  not  known, 
but  the  three  sides  given,  the  following  is  the  rule  which 
the  mere  arithmetician  must  take  on  trust. 

Rule  4.  Add  all  the  three  sides  together,  and  take 
half  that  sum.  Next,  subtract  each  side  severally  from 
the  said  half  sum,  obtaining  three  remainders.  Then 
multiply  the  said  half  sum  and  those  three  remainders 
all  together,  and  extract  the  square  root  of  the  last  pro- 
duct, for  the  area  of  the  triangle. 

To  find  the  area  of  a  trapezoid,  or  a  figure  of  4  sides, 
and  only  two  opposite  sides  parallel. 

Rule  5.  Add  the  two  parcdlel  sides  together,  and 
take  half  the  sum :  this  will  be  the  mean  or  average 
width.  Multiply  this  mean  vndth  by  the  perpendicular 
distance  between  the  parallel  sides> 

To  find  the  area  of  a  circle. 

Rule  6.  Square  the  diameter,  and  multiply  that 
square  by  the  decimal  ,7854. 

Or,  when  most  convenient.  Multiply  half  the  diam- 
eter  by  half  the  circumference. 

[Note.  Tlie  circumference  to  the  diameter  of  any  cir- 
cle, is  found  to  be  very  nearly  in  the  proportion  of  22  to 
7:  a  more  accurate  ratio  is  .3,1416  to  1.] 

To  find  the  surface  of  a  globe  or  sphere. 

Rule  7.  Multiply  the  circumference  and  diameter 
toor.ther  :  or,  inultiply  the  square  of  the  circumference 
6y  3,1416. 


216  ARITHMETIC. 


This  rule  must  be  taken  on  authority.  For  proof,  see 
Geometry. 

Definitions.  A  right  cylinder  is  a  round  body  of 
uniform  diameter  ;  its  two  ends  parallel  and  at  right  an- 
gles with  its  length.  Conceive  the  surface  of  such  a  cy- 
linder cut  lengthwise  and  rolled  out,  it  will  form  a  paral- 
lelogram whose  width  is  the  circumference  of  the  cylin- 
der ;  therefore,  to  find  the  surface  of  a  cylinder,  we  have 
the  following 

Rule  8.  Multiply  the  length  by  the  circumference. 

A  right  cone  may  be  covered  by  a  multitude  of  equal 
isosceles  triangles,  whose  base  is  the  base  of  the  cone, 
and  whose  altitude  is  the  slant  height  of  the  cone ;  there- 
fore, 

To  find  the  surface  of  a  right  cone. 

Rule  9.  Multiply  the  circumference  of  the  base  by 
half  the  slant  height. 

To  find  the  area  of  an  ellipse. 

Rule.  Multiply  the  two  diameters  together,  and  that 
product  by  the  decimal  ,7854. 

EXAMPLES  UNDER  THE  FOREGOING  RULES. 

1.  There  is  a  square  of  80  rods  on  a  side,  and  a  par- 
allelogram also  of  80  rods  each  side  ;  but  from  the  base 
to  the  opposite  side  is  only  70  rods  of  perpendicular  dis- 
tance ;  what  is  the  difference  in  area  of  these  two  figures? 

Jim.  800  rods. 

2.  How  many  yards  in  a  triangle  whose  base  is  148 
feet,  and  perpendicular  45  feet  ?  .^ns.  370  yards. 

3.  The  three  sides  of  a  triangle  are  severally  60,  50, 
and  40  rods  ;  how  many  acres  does  it  contain  ? 

^ns.  61. 

4.  On  a  base  of  120  rods,  a  surveyor  wished  to  lay  off 
a  rectangular  lot  of  land  containing  40  acres  ;  what  dis- 
tance in  rods  must  he  run  from  his  base  line  ? 

Jlns.  5S^  rods. 

5.  How  many  boards  will  it  require  to  cover  the  body 
of  a  barn  60  feet  long,  40  feet  wide  and  20  feet  high,  to 


MENSURATION.  217 


the  eaves,  the  boards  being  on  an  average  15  inches  wide 
and  10  feet  long?  Ans.  .S20  hoards. 

6.  How  many  feet  in  a  board  22  inches  wide  at  one 
end,  8  inches  wide  at  the  other,  and  14  feet  long? 

^ns.  17^  feet. 

7.  A  board  is  9  indies  wide ;  how  much  in  length 
will  it  require  to  make  6|  square  feet?     .'?».s-.  8|  feet. 

8.  The  bottom  of  a  circular  cistern  is  5^  feet  in  diam- 
eter ;  how  many  square  feet  of  surface  ? 

.^ns.  23,758+ feet. 

9.  What  quantity  of  surface  in  a  cylinder  30  feet  long 
and  3  feet  in  diameter,  the  two  ends  included  ? 

£ns.  296,88-}-feet. 

10.  A  man  bought  a  farm  198  rods  long,  and  150  rods 
wide,  and  agreed  to  give  $32  per  acre  ;  what  did  the  farm 
come  to  ?  ^ns.  $5940. 

N.  B.  Make  no  attempt  to  compute  the  number  of 
acres  definitely. 

11.  If  the   forward  wheels  of  a  coach  are  4  feet,  and  j 
the  hind  ones  5  feet  in  diameter,  how  many  more  times 
will  the  former  revolve  than  the  latter,  in  going  a  mile  ? 

.dns.  84. 
N.  B.  In  this  problem  use  7  to  22. 

The  solutions  of  the  foregoing  problems  are  all  very 
brief  by  canceling,  except  ihe  3d  and  9th. 

(Art,  121.)  The  foregaing  rules  apply  to  Plasterers^ 
Pavers^  and  Carpenters^  loork.  These  artificers  sfener- 
ally  compute  their  work  at  so  much  a  square  yard,  or^o 
much  for  100  square  feet,  which  last  is  sometimes  called 
a  carpenter's  square. 

To  compute  the  number  of  square  feet,  yards,  &;c.,  the 
length,  breadth  and  heighth  of  a  room  is  taken  in  feet  and 
inclies,  never  measuring  closer  than  inches.  In  books, 
we  often  meet  with  feet,  inches,  thirds,  fourths,  <fec.,  de- 
scending on  a  scale  of  12,  called  duodecimals,  the  foot 
being  the  unit ;  hence  inches  are  fractions,  and  two  frac- 
tions multiplied  together,  that  is,  taking  part  of  a  part, 
makes  a  less  part.  Tiierefore,  inches  multiplied  by  indi- 
es, must  give  less  than  inches,  that  is,  give  thirds,  &c. 


218  ARITHMETIC. 


As  a  study  of  numbers,  a  few  examples  in  duodecimals, 
might  be  of  some  advantage  ;  but  with  them  we  must  re- 
tain cumberous  and  clumsy  modes  of  operations,  which 
are  of  no  practical  value,  and  thus  cherish  bad  habits  that 
ought  to  be  forgotten  ;  v:e  shall ^  therefore,  dispense  with 
duodecimals.  AH  practical  men  take  no  account  of  meas- 
ures lower  than  inches. 

EXAMPLES. 

[Note.  In  the  solution  of  problems,  we  always  call 
inches  part  of  a  foot;  6  inches  is  !,,  7  inches  y^,  9  inches 
\,  &c.] 

1.  What  will  it  cost  to  pave  a  foot-path  40  feet  long, 
and  7V  feet  wide,  at  22  cents  per  square  yard  ? 


Solution.  2 

9 
100 


40 
15 
22 


Ans.  7^  dollars. 


2.  What  will  it  cost  to  paint  a  room  which  meas- 
ures 82  feet  6  inches  in  compass  and  12  feet  9  inches 
high,  at  16  cents  a  square  yard?  Ans.  $18,70. 

3.  A  floor  is  36  feet  4  inches,  by  14  feet  9  inclies  ; 
what  will  it  cost  to  lay  it  at  3  dollars  2.5  cents  per  square  ? 

Ans.  $17,42  nearly. 

4.  What  will  it  cost  to  roof  a  building  42  feet  long, 
and  the  rafters  on  each  side  being  17  feet  6  inches  long, 
at  3  dollars  and  25  cents  per  square  of  100  feet  ? 

Ans.  $47,771. 

5.  What  will  the  plastering  of  a  ceiling  come  to,  at  15 
cents  per  yard,  it  being  21  feet  7  inches  long,  and  12  feet 
6  inches  broad  ?  Ans.  $4,47  -\- 

6.  What  will  the  whitewashing  of  a  room  cost  at  2 
cents  per  yard,  allowing  it  to  be  30  feet  6  inches  long,  24 
feet  9  inches  wide,  and  10  feet  6  inches  high,  no  deduc- 
tions beino-  made  for  vacuities  ?  Ans.  $4,23  -}- 

7.  A  room  is  20  feet  6  inches  long,  and  13  feet  6  inch- 
es in  width,  and  10  feet  high ;  what  will  it  cost  to  paper 
the  walls,  at  27  cents  per  square  yard,  deducting  a  fire- 
place of  4  feet  4  inches,  and  3  windows,  each  6  feet  by 
3  feet  2  inches?  Ans.  $18,17. 


MENSURATION    OF    SOLIDS,  ETC.  219 


8.  A  room  is  27  feet  6  inches  by  22  feet  6  inches,  and 
10  feet  3  inches  from  the  wash-board  up.  In  said  room 
is  a  lire-place  of  4  feet  by  4  feet  6  inches ;  2  doors,  each 
8  feet  by  4  feet  4  inches,  and  2  windows,  each  6  feet  by 
3  feet  4  inches.  Wiiat  will  it  cost  to  plaster  said  room, 
at  18  cents  per  yard,  sides  and  ceiling,  deducting  for  va- 
cuities '?  Ans.  $30,32. 

9.  What  is  the  cost  of  smoothing  and  polishing  a  mar- 
ble slab,  whose  length  is  5  feet  7  inches,  breadth  2  feet  3 
inches,  at  38  cents  per  square  foot  ?        Ans.  $4,52|. 


MENSURATION  OF  SOLIDS  AND  CAPACITIES. 

(Art.  122.)  It  will  be  recollected  that  surfaces  are 
measured  by  squares  (Art.  118):  so  solids  are  measured 
by  unit-cubes,  whatever  size  cube  we  take  for  the  unit. 
A  nihe  is  a  solid  figure  of  six  equal  square  surfaces, 
and  all  its  angles  right  angles.  To  measure  a  riglit 
angular  solid,  such  as  a  square  stick  of  timber,  for  in- 
stance, tlie  surface  of  the  end  is  measured  by  little  squares ; 
and  if  we  saw  off  a  length  equal  to  a  side  of  one  of  tliese 
little  squares,  it  is  plain  that  we  shall  have  as  many  cubes 
as  the  surface  of  the  end  contains  squares;  twice  this 
length  will  give  twice  as  many  cubes ;  three  times  the 
length,  three  times  as  many  cubes,  &;c.  &c.  But  to  find 
the  number  of  squares  in  the  end,  we  multiply  the  width 
and  depth  of  the  end  together ;  then  that  product  by  th.e 
length,  and  we  have  the  number  of  unit-cubes  in  the 
whole  solid.  All  solid  figures,  of  whatever  shape,  are 
referred  to  regular  right  angular  solids,  directly  or  in- 
directly. 

To  reduce  one  solid  figure  to  its  equivalent,  in  another 
shape,  is  the  office  and  object  of  solid  geometry,  not 
arithmetic;  hence  our  rules  are  drawn  from  geometry; 
and,  as  arithmeticians,  we  must  take  them  on  authority. 
We  must  advance  to  geometry  to  perfectly  comprehend 
them. 


220  ARITHMETIC. 


BefinUion  1st.  To  find  the  cubical  contents  of  a  right 
angular  solid. 

Rule  1.  JMuUipJij  length,  breadth  and  depth  together. 

[Note.  No  school  should  be  without  a  set  of  geomet- 
rical solids.  A  bare  inspection  of  them  will  give  a  bet- 
ter idea  of  the  regular  solids,  than  the  most  exact  defini- 
tion, or  the  most  minute  description.] 

Bef,  2.  A  prism  is  a  solid,  whose  ends  are  any  simi- 
lar equal  and  parallel  surfaces,  and  its  sides  rectangles ; 
hence,  to  find  the  solidity  of  a  prism, 

Rule  2.  Multiply  the  area  of  one  end  by  the  length. 

Def.  3.  A  cylinder  is  a  circular  solid,  equal  at  both 
ends.  A  right  cylinder  has  circular  ends,  at  right  angles 
to  the  axis.     To  find  the  contents  of  such  a  cylinder. 

RtTLE  8.  Multiply  the  area  of  one  of  the  circular  ends 
by  the  length. 

To  estimate  the  contents  of  a  wedge  formed  solid. 

Rule  4.  Multiply  the  area  of  its  end  by  half  of  its 
perpmdicidar  length. 

Def.  4.  A  pyramid  is  a  solid  which  decreases  uniform- 
ly from  its  base,  until  it  comes  to  a  point.  The  base  of 
a  pyramid  may  be  either  a  square,  a  triangle,  or  a  circle. 
Hence  it  is  named  a  square  pyramid,  a  triangular  pyra- 
mid, or  a  cone.  The  point  in  which  the  pyramid  ends,  is 
called  its  vertex;  a  straight  line  from  the  vertex  at  right 
angles  to  the  base,  is  called  the  altitude.  To  find  the 
solidity  of  a  pyramid  or  cone. 

Rule  5.  Multiply  the  area  of  the  base  by  one-third 
of  the  altitude. 

Def.  5.  When  the  smaller  end  of  a  pyramid  or  cone 
is  cut  off  parallel  to  its  base,  the  residue  is  called  the  frus- 
trum  of  a  pyramid  or  cone.  When  the  base  is  a  square, 
the  following  rule  will  give  the  solidity  of  a  frustrum : 

Rule  6.  Square  both  upper  and  lower  base :  and,  to 
flu  .nnn  of  these  squares,  add  the  product  of  the  sides 


MENSURATION    OF    SOLIDS,  ETC.  221 


of  upper  and  lower  bane,  and  midtiphj  this  last  sum  by 
one-third  of  the  altifiide  of  the  frutitrum. 

When  the  base  is  a  circle. 

Rule  7.  To  the  sum  of  the  squares  of  the  two  diam- 
eters, add  their  product ;  multiply  this  sum  by  one-third 
of  the  altitude,  and  this  last  product  by  the  decimal 
,7854. 

To  find  the  solidity  of  a  sphere  or  globe. 

Rule  8.  Find  the  surface  by  rule  7,  Art.  120,  then 
multiply  the  surface  by  \  of  the  diameter ;  or,  which 
is  the  same  thing;  multiply  the  square  of  the  diameter 
by  the  circumference,  and  divide  the  product  by  6.  Or, 
we  may  cube  the  diameter,  and  multiply  it  by  the  deci- 
mal ,5236.  Or,  if  more  convenient,  cube  the  circum- 
ference, and  midtiply  it  by  ,01688. 

To  find  the  solid  contents  of  a  spherical  segment. 

Rule  9.  From  three  times  the  diameter  of  the  sphere, 
take  double  the  height  of  the  segment ;  flien  multiply 
the  remainder  by  the  square  of  the  height,  and  that 
product  by  the  decimal  ,5236,  and  ive  have  the  content. 

Tlie  foregoing  rules  have  a  variety  of  practical  ap- 
plications to  business  and  the  mechanic  arts.  Irregular 
bodies  may  be  measured  approximately,  by  imagining 
them  cut  and  divided  so  that  the  parts  in  succession  may 
make  some  one,  two,  or  more  of  the  mathematical  figures  : 
this  is  done  by  engineers  in  measuring  excavations  and 
embankments.  We  can  measure  the  cubical  contents  of 
the  most  irregular  stone  in  the  field,  by  putting  it  into  a 
regular  vessel  of  water,  and  observing  the  cubical  contents 
of  the  rise  f  or,  by  taking  the  specific  gravity  of  a  body, 
we  can  estimate  its  contents  from  its  weight;  but  this 
cames  us  out  of  arithmetic  into  philosophy. 

QUESTIONS    UNDER    THE    FOREGOING    RULES. 

1.  How  many  solid  feet  in  a  block  6  feet  long,  4  feet 
high,  and  3  wide  ?     Solution  :  6X  4X  3=72  feet,  Ans. 

*  A  boy  once  made  a  wager  with  one  of  his  companions,  that  he 
could  measure  the  cubical  contents  of  a  certain  pile  of  brush.  He 
did  so  by  cutting  it  short,  and  sinking  it  m  a  regular  tan-vat. 

"  ~^       f2  ~~~ 


222 


ARITHMETIC. 


2.  How  many  cubic  feet  in  a  piece  of  timber  18  feet 
long,  1  foot  6  inches  high,  and  1  foot  3  inches  wide  ? 

.^ns.  33  ^ 


Sohition : 


8 

18 

U 

or, 

2 

3 

n 

4 

5 

12 

40 

12 

12 

12 

U 

8 

3.  Tlie  plate  supporting  the  rafters  of  a  house  being 
40  feet  long,  14  inches  wide,  and  8  inches  thick,  how 
many  solid  feet  does  it  contain  1  Ans.  31^  feet. 

Take  this  form  for  all  solids,  Art.  24. 


inches. 


N.  B.  To  find  how  many  cords  in  a  given  pile  of  wood, 
to  find  how  many  bushels,  gallons,  &c.,  in  any  given  ca- 
pacity, we  must,  of  course,  divide  the  whole  contents  ex- 
pressed in  cubic  feet,  or  inches,  as  the  case  may  require, 
by  the  feet,  or  inches,  &c.,  which  make  one  cord,  or  one 
bushel,  &;c.,  of  the  required  dimensions.  As  an  illustra- 
tration,  observe  the  solution  of  the  following  example. 
The  dimensions  must  be  reduced  to  cubic  inches. 

3.  There  is  a  rectangular  cistern,  whose  length  is  14 
feet,  depth  16  feet,  and  breadth  9  feet;  how  many  bar- 
rels will  it  hold,  allowing  231  cubic  inches  to  a  gallon, 


and  32  gallons  to  a  barrel  ? 

Solution : 

Divisors, 


{ 


231 


32 


Ans.  471yV 
inches  in  length. 

inches  in  depth. 

inches  in  width. 


This  cancels  down  to  11,  for  a  divisor,  and  3  times  1728 
for  a  dividend ;  which  quotient  gives  the  answer. 

4.  A  block  of  marble  is  6  feet  long,  3  feet  wide,  and 
1  foot  4  inches  thick ;  how  much  will  it  cost,  at  30  cents 
per  cubic  foot?  Ans.  $7,20. 

5.  If  a  pile  of  wood  be  60  feet  long,  12  feet  high,  and 
6  feet  wide,  how  many  cords  does  it  contain  ? 

Ans.  33?  cords. 


MENSURATION    OF    SOLIDS,  ETC.  223 


6.  The  bin  of  a  granary  is  10  feet  long,  5  feet  wide,  and 
4  feet  high  ;  allowing  the  cubical  contents  of  a  dry  gal- 
lon to  be  2G8j  cubic  inches;  how  many  bushels  of  grain 
will  it  contain  1  Ans.  160,^. 

7.  If  you  wanted  a  bin  to  contain  twice  as  much  as 
mentioned  in  the  last  problem,  with  a  lengdi  of  12  feet, 
and  a  breadth  of  6  feet,  of  what  height  must  it  be  ? 

Ans.  ^l  feet. 

8.  A  canal  contractor  engaged  to  excavate  2  miles  of 
canal  across  a  plane,  at  8  cents  per  cubic  yard  :  the  canal 
to  be  54  feet  wide  at  top,  40  at  bottom,  and  4%  feet  deep; 
what  did  it  amount  to  ?  Ans.  $6617,60. 

9.  What  is  the  area  of  a  circle  of  5  feet  in  diameter? 

Ans.  19,635  feet. 
10.  What  is  the  area  of  a  circle,  whose  diameter  is  5 
feet  6  inches  ?  Ans.  23,758-f 

Remember,  that  similar  figures  are  to  each  other  as  the 
squares  of  their  like  dimensions  (Art.  120).  Apply  this 
principle  to  problem  10,  and  we  have,  as  (5)^  :  to  (S't)^; 
or  as  10-:  U^.  That  is,  as  100  :  121  :  19,635  : 
23,75835. 

11.  There  is  a  circular  cistern,  of  uniform  diameter, 
whose  depth  is  8  feet,  and  diameter  5  feet ;  what  is  its 
capacity,  allowing  231  inches  to  the  gallon,  and  how 
much  would  its  capacity  be  increased  by  adding  6  inches 
to  its  diameter?      ^        C  1175,04  gallons  its  capacity. 

\      246^-  gallons  increase. 

12.  A  man  wishes  to  make  a  cistern,  of  8  feet  in  di- 
ameter, to  contain  60  barrels,  at  32  gallons  each.,  \\m\  231 
cubic  inches  to  a  gallon ;  what  shall  be  tlic  deptli  of  the 
cistern  ? 

60X32X231  gives  the  cubic  inches  the  cistern  i:;  to 
contain.  This,  divided  by  the  circular  end,  expressed  in 
inches,  will  give  the  depth  in  inches.  Ans.  61  J-  inches. 

13.  A  stick  of  timber  is  8  inches  wide,  and  5  inches 
thick ;  how  much  in  length  will  it  require  to  make  3  cu- 
bic feet  ? 

N.  B.  The  principles  contained  in  Rule  2  will  solve 
this  problem. 


224  ARITHMETIC. 


Solution, 

2  ^ 
5 


3 

i^  3 

12 

^^         Ans.  104  U%i. 

14.  How  many  bricks  will  it  require  to  build  a  wall 
44  feet  long,  42  I'eet  wide,  (on  the  outbide,)  and  25  feet 
high,  and  the  wall  to  be  1  foot  thick,  allowing  20  bricks  to 
the  solid  foot  ?  Ans.  84000. 

15.  What  will  it  cost  to  build  a  brick  wall  240  feet 
long,  6  feet  high,  and  3  feet  thick,  at  3  dollars  25  cents 
per  1000  bricks — each  brick  being  9  inches  long,  4  inches 
wide,  and  2  inches  thick  ?  Ans.  $336,96. 

16.  A  ship's  hold  is  lb\  feet  long,  18.V  feet  wide,  and 
1\  deep;  how  many  bales  of  goods  3",  feet  long,  2^^  feet 
deep,  and  2^  wide,  may  be  stowed  therein,  leaving  a 
gangway,  the  whole  length,  of  3^  feet  wide? 

Ans.  385,4  + 
N.  B.  Do  this  by  one  operation ;  after  taking  out  the 
gangway  mentally ^  by  subtraction. 

17.  A  stick  of  timber  is  16  inches  broad  and  8  inches 
thick  ;  how  many  feet  in  length  must  be  taken  to  make 
20  solid  feet?  Ans.  221. 

18.  How  many  solid  feet  in  a  stick  of  timber  30  feet 
long,  16  inches  broad,  and  9  inches  thick? 

Ans.  30  feet. 

19.  How  many  cubic  feet  in  a  piece  of  timber  19  feet 
6  inches  long,  1  foot  6  inches  broad,  and  I  foot  4  inches 
thick  ?  Ans.  39  cubic  feet. 

20.  There  is  a  square  pyramid,  each  side  of  whose 
base  is  30  inches,  and  whose  perpendicular  height  is  120 
inches,  to  be  divided  by  sections,  parallel  to  its  base,  into 
three  equal  parts;  required  the  perpendicular  heigiit  of 
each  part. 

Ans.  The  height  of  the  lower  section  is  15,2  inches; 
the  height  of  the  middle  section  is  21,6  inches;  the 
height  of  the  top  section  is  83,2  inches. 

N.  B.  It  is  proved  in  geometry,  that  similar  solids  are 
to  one  another  as  the  cubes  of  their  like  dimensions. 
This  principle  solves  problem  20. 


MENSURATION    OF    SOLIDS,  ETC.  225 


21.  The  base  of  a  cone  is  8  feet  in  diameter,  and  60 
feet  high  ;  how  many  cubic  feet  does  it  contain  ? 

.^iu\  1005,6. 

22.  A  stick  of  round  timlier  is  2  feet  in  diameter  at 
one  end,  and  10  inches  at  the  other,  and  20  feet  long; 
how  many  solid  leel  does  it  contain  ? 

Solution,  Ihth  7. 

(24--fl0=^+240)20X12X,7854 

^ ' =22,83+feet,  Arts. 

3.12.12.12 

Rule  7th  can  be  apph'ed  to  gai/gino;,  or  measuring  of 
casks.  A  common  cask,  of  circular  form,  larger  in  di- 
ameter in  the  middle  than  at  the  ends,  may  be  considered 
as  composed  of  two  equal  fntshims  of  a  cone,  and  its 
contents  found  in  inches  accordingly;  which  contents, 
divided  by  231,  for  wine  gallons,  282  for  beer  gallons. 
For  bushels,  dividedby  2 150,4?  or,  for  all  practical  purpo- 
ses, by  2150  is  sufficiently  accurate.  A  bhshel  of  corn, 
lime,  &c.,  is  2688  inches. 

EXAMrLES. 

1.  The  bung-diameter  of  a  cask  is  38  inches,  the  head- 
diameters  inside  the  staves  28  inches,  and  the  length  45 
inches  ;  how  many  wine  gallons  will  it  contain  ? 

Jins.  107,89-1- 

(38--|-28--|-38.28),7854  X  45 

Solution,  •   . 

3X231 

Now  observe,  that  the  decimal  ,7854,  and  the  divisor 
231,  for  wine  gallons,  will  appear  in  any  problem  ;  and 
as  the  decimal  is  divisible  by  231,  and  give  the  quotient 
,0034,  we  may  give  the  following  abridged  ride  for  Jin  d- 
ing  the  wine  gallons  in  a  cask. 

Rule.  To  the  square  of  the  head  and  hung-diame- 
fers,  add  their  product ;  multiphj  the  sum  by  i  of  the 
l&ngth,  and  that  product  by  the  decimal  ,0034. 

2.  The  head-diameter  of  a  cask  is  16  inches,  the  bung- 
diameter  18  inches,  and  the  length  27  inches;  what  is 
the  content  of  die  cask  in  wine  gallons  ?    Ans.  20,56. 

3.  In  a  circular  vessel,  whose  greater  diameter  is  80 


226  ARITHMETIC. 


inches,  the  less  71,  and   the  depth  34,  what  is  the  con- 
tents in  liquid  gallons,  and  also  in  bushels  of  grain  ? 

^         5  659,72+ gallons. 
*^"'^^*    I    70,86  bushels. 

4.  The  greater  diameter  of  a  tub  is  38  inches,  the  less 
20,2,  and  the  depth  21  ;  M^hat  is  the  content  in  gallons  ? 

,.^ns.  62,34+  gallons. 

5.  The  top  diameter  of  a  tube  is  22  inches,  the  bottom 
40,  and  the  height  60  ;  what  is  its  contents  in  gallons, 
also  in  bushels  of  grain  ?  /?       S  201,55+  gallons. 

'^^^^'  I    21,64+  bushels. 

6.  How  many  barrels,  of  32  gallons  each,  in  a  cistern 
whose  greater  diameter  is  8  feet  6  inches,  less  8  feet,  and 
depth  7  feet  9  inches  ?     ^'^ns.  96  barrels  28+  gallons. 

7.  How  many  bushels  of  grain  in  a  bin  that  is  8  feet 
long,  4  feet  wide,  and  6  feet  high  ? 

^  .^ns.  123+  bushels. 

8.  How  many  bushels  in  a  loaded  boat  60  feet  long, 
16  feet  wide,  and  4\  feet  deep  ?    j^ns.  2777+  bushels. 

9.  A  man  bought  a  grindstone,  which  was  48  inches 
in  diameter,  and  5  inches  in  thickness,  for  10  dollars. 
When  he  had  ground  ofl"  3  inches  from  its  circumference, 
his  neighbor  proposed  to  purchase  it  from  him,  giving  a 
proportional  part  of  the  first  cost  for  the  residue  ;  deduct- 
ing 4  inches  every  way  from  the  center,  allowed  to  be 
the  limit  to  which  it  could  be  used.  What  should  he 
have  paid  ■  ^ns.  $7,58. 

10.  Find  the  solid  contents  of  a  sphere,  whose  diame- 
ter is  12,  by  Rule  8  ?  .^ns.  904,78  + 

11.  What  is  the  sohd  contents  of  a  globe  48  inches  in 
diameter  ?  Ans.  64  X  904,78+  inches. 

(Art.  123.)  Connected  with  mensuration,  as  we  have 
already  observed,  are  square  and  cube  figures,  the  sides 
of  which  represent  square  and  cube  roots.  But  to  inves- 
tigate these,  we  must  commence  with  the  general 

INVOLUTION  OF  POWERS. 

A  POWER  is  the  product  of  a  number  multiplied  into 
itself  any  number  of  times. 

The  number  itself  is  called  its  1st  power. 


INVOLUTION"    OF    I'OWERS.  227 

The  number,  multiplied  into  itself,  is  called  the  2nd 
power. 

The  2nd  power,  multiplied  again  by  tlie  number,  is 
called  the  3rd  power. 

The  3rd  power,  again  multiplied  by  tlie  number,  is 
called  the  4th  poM'er,  &c.  &c. 

The  number  so  multiplied  into  itself,  is  called  a  root ; 
and  the  number  of  times  a  root  is  taken  as  a  factor,  de- 
notes its  power.     Thus, 

3=  the  root  or  first  power. 
3X3=3'=9  the  square  or  2nd  power  of  3. 
3X3X3=3^=27  the  cube  or  3rd  power  of  3. 

This  is  as  far  as  numericals,  with  their  roots  and  pow- 
ers, can  be  made  to  correspond  to  geometrical  figures. 
The  following  are  entirely  numerical. 

3X3X3X3=3^=81   the  4th  power  of  3. 
3X3X3X3X3=3^=243  the  5th  power  of  3,  &c. 

The  expression  3''  is  simply  an  abbreviation  of  3  X  3  X 
3X3;  the  small  figure  4  above  and  on  the  right  of  the  3, 
is  called  an  index  or  exponent,  showing  how  many  times 
the  root  is  taken  as  a  factor. 

If  we  take  the  powers  indicated  by  any  two  exponents, 
and  multiply  them  into  each  other,  their  product  will  be 
the  power  indicated  by  the  sum  of  the  two  exponents. 
Thus,  3^X3''=3^  the  exponent  of  which  is  ^;  or,  if  we 
multiply  any  power  into  itself,  the  product  will  be  the 
power  indicated  by  the  addition  of  the  exponent  to  itself. 
Thus,  8^X  8=^=8^  Likewise,  if  a  higher  power  be  divi- 
ded by  a  lower,  the  quotient  will  be  the  power  indicated 
by  the  difl^erence  of  the  exponents.  Thus,  16'^-^16''= 
16^.  Therefore,  to  multiply  powers  of  the  same  root,  we 
simply  add  the  indices,  and  to  divide,  we  subtract  the  in- 
dices or  exponents. 

JVeJind  the  power  of  any  number  by  multiplication. 

EXAMPLES. 

1.  What  is  the  3rd  power  of  2  ?  Ans.  8. 

2-  What  is  the  5th  power  of  4  ?  Ans.  1024. 

3.  What  is  the  6th  power  of  8  ?  Ans.  262144. 


228  ARITHMETIC. 


Problem  third  is  performed  thus  :  8''=512  ;  but  S^X 
83=8^     Hence  512X512=262144.^^15. 

4.  Find  the  square  of  56.  Jins.  3136. 

5.  Find  the  cube,  or  3rd  power,  of  67. 

Ans.  300763. 

6.  What  is  the  cube  of  186  ?  Ans.  6434856. 

7.  What  is  the  3rd  power  of  9  ?  Ans.  729. 

8.  What  m  the  9th  power  of  9  ? 

Am.  729X729X729=387420489. 

9.  What  is  the  7th  power  of  9  ? 

Ans.  729X729X9=4782969. 
10.   What  is  the  3rd  power  of  1?  Ans.  1.     The  4th 
power  of  1  ?  Ans.  1.     The  10th  power  of  1  ?     Ans.  1. 
Any  power  of  1  ?  Ans.  1. 

(Art.  124.)  By  the  nature  of  multiplication  and  the 
definition  of  powers,  the  pupil  will  perceive  that  any 
power  or  root  of  1,  is  neither  more  nor  less  than  1,  for 
the  multiplication  of  one  into  itself  any  number  of  times, 
cannot  change  it;  but  all  numbers  greater  than  one,  gives  | 
higher  numbers  for  powers,  and  the  higher  the  power,  \ 
the  higher  the  number  above  its  root.  1 

We  now  propose  to  examine  the  powers  of  a  proper  j 
fraction.     Suppose  it  required  to  find  the  square  or  se-  \ 
cond  power  of  ~  ;  that  is,  to  multiply  |-  by  l ;  or,  to  take  I 
1,  one-thiid  of  one   time,  which   must  be  ^th  the  power  i 
less  than  its  root.    Now  it  must  be  very  clear  to  all  who  ; 
will  take  the  trouble  to  call  it  to  mind,  that  a  fraction  mul- 
tiplied  into  itself,  is  taking  a  part  of  a  part,  ivhich  must 
be  still  less  ;  hence  any  power  of  a  fraction  must  be  less, 
and  of  course,  the  reverse,  any  root  of  a  fraction  must 
be  greater  than  the  power.    We  can  further  illustrate  this 
by  fig.  1,  Art.  118.     If  we  take  the  whole  line  for  unity, 
the  area  of  the  figure  is  unity,  and  one  section  will  re- 
present 1,  and  its  square  is  i-  of  the  whole,  as  we  may 
see  by  inspection. 

EXAMPLES. 

1.  Find  the  cube  of  \.  Ans.  \. 

2.  What  is  the  3rd  power  of  ^  ?  Ans.  ■^\^. 

3.  Find  the  4th  power  of  ,3.  Ans.  ,0081. 

4.  Find  the  4th  power  of  ,05.  Ans.  ,00000625. 


EVOLUTION,  ETC. 


229 


5.  AVhat  is  the  2nd  power  of  |  ? 

6.  What  is  the  3d  power  of  f  ? 

7.  What  is  the  5th  power  of  ^  ? 

8.  What  is  the  cube  of  ,25  ? 

9.  What  is  the  square  of  ,75  ? 
10.  What  is  the  cube  of  li  ? 


Ans.  i|A. 

^;i.s.  ,015625. 
Jlns.  ,5625. 


EVOLUTION,  OR  THE  EXTRACTION  OF 
ROOTS. 

(Art.  125.)  Evolution  is  the  reverse  of  Involution. 
In  a  practical  point  of  view,  it  is  only  important  that  we 
should  learn  how  to  extract  two  roots,  the  square  and  the 
cube  roof.  In  the  higher  mathematics,  we  extract  the 
roots  of  higher  powers  by  logarithms;  and,  in  fact,  math- 
ematicians generally  extract  all  roots  by  means  of  loga- 
ritlims  ;  but  it  is  important,  on  many  accounts,  that  we 
learn  the  common  method  of  extraction. 

As  evolution  is  the  reverse  of  involution,  we  must  care- 
fully notice  how  roots  form  powers  by  multiplication,  that 
we  may  trace  back  the  operation  from  powers  to  their 
roots.  As  a  preliminary  step,  we  give  the  following  ta- 
ble with  the  first  10  numbers,  with  their  corresponding 
squares  and  cubes,  which  the  pupil  should  so  familiarize 
to  his  eye,  as  to  recognize  them  as  square  or  cube  num- 
bers, as  the  case  may  be,  as  soon  as  seen. 


Numbers, 

1 

2 

3   4 

1 

5 

6 

7 

8 

9 

10 

Squares, 

1 

4 

9 

16 

25 

36 

49 

64 

81 

100  i 

■Cubes. 

1 

8  27  64 

1 

125 

216 

343 

512 

729 

1000 

Any  number  will  produce  a  perfect  power,  by  involu- 
tion ;  yet  there  are  many  numbers  that  huve  no  perfect 
roots.  Imperfect    roots  are  called  surds,    and  such 

can  only  be  found  approximately,  but  to  a  sufficient  de- 
gree of  exactness  for  all  priictical  utility.  For  example, 
36  and  49  are  square  numbers,  and  all  the  other  whole, 
and  many  mixed  numbers  between  36  and  49,  have  surd 


230  ARITH3IETIC. 


roots.  Very  few  numbers  are  perfect  powers  of  more 
than  one  root;  thus,  8  is  a  cube  number,  but  not  a 
sqvare  number,  &;c. 

(Art.  126.)  As  a  very  important  and  useful  fact,  we 
call  the  pupil  to  notice,  that  any  root  composed  of  one 
Jigure,  its  square  may  be  composed  of  two  figures,  and 
no  more  than  two;  and  its  cube  may  be  composed  of 
three  figures,  and  no  more  than  three,  as  will  be  seen  by 
inspecting  the  preceding  table. 

Again  :  every  root  composed  of  two  figures,  may  have 
more  than  2,  and  not  more  than  4  figures,  in  its  square  ; 
and  may  have  more  than  3,  and  not  more  than  6  figures 
in  its  cube,  as  we  may  learn  by  involution ;  thus. 
The  lowest  root  of  2  orders,  is  10,  the  highest  is  99 

Squares, 100 9801 

Cubes, 1000 970299 

Lowest  root  of  3  places,  is  •  -100,  highest  is  .  .  999 

Squares. 10000 998001 

Cubes, 1000000  ....  997002999 

Hence,  for  every  figure  in  a  root,  we  may  have  two  in  its 
square  and  three  in  its  cube.  This  fact  teaches  us,  in 
preparing  to  extract  the  square  root,  to  separate  the  pow- 
er off  inio  periods  oUwo  figures  each,  and  in  preparing 
to  extract  the  cube  root,  to  separate  the  power  off  into 
periods  of  ^figures  each,  comynencing  at  the  unit.  The 
number  of  periods,  then,  in  both  cases,  will  indicate  the 
number  of  figures  in  the  root. 

Now  let  us  turn  exclusive  attention  to 

SQUARE  ROOT. 

(Art.  127.)  There  are  two  ways  of  explaining  the 
principle  of  the  operation,  in  extracting  square  root:  one 
by  retracing  the  numerical  operation  of  squaring,  the  oth- 
er by  referring  and  comparing  the  power  to  a  geometrical 
square  surface.  Both  of  these  should  be  understood. 
To  prepare  the  way  for  a  clear  understanding  of  this  im- 
portant branch  of  arithmetic,  the  pupil  should  square  sev- 
eral numbers,  consisting  of  two  figures,  and  scrutinize  ev- 
ery step  of  the  operation :  thus, 


SQUARE    ROOT.  231 


Let  it  be  required  to  square 35 

Write  the  number  under  itself, 35 

Tlie  units  by  the  units,  give 25 

The  units  below  by  tens  above,  give  •  •  •  .150 
The  tens  below  by  units  above,  give  •  •  .150 
The  tens  by  the  tens  give  liundreds,    •   .  900 

Product  or  square, 1225 

Now  observe,  that  this  square  is  made  up  of  the  square 
of  the  tens,  twice  the  product  of  the  tens  into  the  units, 
and  the  square  of  the  units.  We  must  now  suppose  our- 
selves ignorant  of  the  real  root,  35,  and  by  the  above 
principle,  find  it.  First,  divide  the  power  ofl"  into  pe- 
riods of  two  figures  each  (Art.  126),  and  it  will  stand  thus : 

1215. 
Being  two  periods,  there  will  be  two  figures  in  the  root, 
a  unit  and  a  ten.     Now,  the  square  of  the  tens  can  form 
no  part  of  the  Knit  period,  as  the  square  of  one  10  pro- 
duces  100,  and    the   square  of  tens    generally  produce 
hundreds,  and  nothing  less.      Hence,  the  greatest  square 
contained  in  12,  the  2d  period,  is  the  square  of  the  tens  ; 
but  9  is  the  greatest  square  number  less  than  12,  and  3 
its  root;  and  so  far,  the  operation  may  stand  thus  : 
1225(3 
9 


325 

But  the  tens  and  units  are  twice  multiplied  together, 
and  their  product  forms  no  part  of  the  units  in  the  pow- 
er; hence,  if  we  divide  the  32  tens,  by  double  the  tens 
in  tlie  root,  we  must  have  the  other  factor,  or  the  units  of 
the  loot.     Now  the  operation  stands  thus  : 

12*25(35 
9 


Double  of  3  is  6  65  325 
6  in  32,  5  times,  325 
5  times  65  is  325,  


000 


232 


ARITHMETIC. 


120 

9 

H 

X600 

I 

120 

(Art.  128.)  Let  us  now  illustrate  the  mode  of  extrac- 
tion, by  means  of  a  geometrical  square.  We  cannot  de- 
mand the  square  root  of  any. number,  without  first  sup- 
posing it  to  represent  the  area  of  some  square  surface, 
and  tlie  root  is  the  length  of  a  side  of  the  supposed 
square  ;  but  to  brina:  it  before  the  mind  in  a  definite  man- 
ner, we  demand  the  square  root  of  1849.  This  number, 
then,  must  represent  a  square. 

Suppose    the    figure       A  E         B 

ABCD  contains  1849 
square  feet,  and  that  the 
number  consists  of  two 
periods  ;  then  there  must 
be  2  figures  in  the  root. 
The  largest  root  whose 
square  can  be  taken  out 
of  the  left  hand  period, 
is  4,  (or,  as  it  will  stand 
in  ten's  place  in  the 
root,  it  is  40,)  and  the  D 
square  of  this  is  16,  (or 
1600.)  This  taken  from 
the  whole  square  ABCD 
or  1849,  leaves  249. 

Now  double  GH,  or 
HI,  which  is  40,  for  a 
divisor,  omitting  the  ci^ 
pher  to  leave  place  for 
the  next  quotient  figure, 
to  complete  the  divisor, 

80  into  249  is  con- 
tained 3  times  ;  this  3  is 
the  widtli  of  the  oblong 
AEHG,  orHFCI.  But 
the  square  is  imperfect 
without  EBFH;  then 
annex  the  three  to  the 
divisor.  Now  multiply 
this  perfect  divisor  by 
the  last  fiffure  of  the  root, 


18,49(43 
16 

83)249 
249 


EHID  =1600 
AEHG  =  120 
HFCl  ==■  120 
EBFH  =       9 


ABCD  =1819 


SQUARE    ROOT.  233 


to  get  the  quantity  in  the  two  oblontr  (lornrcs,  and  the  small 
square  whicli  comprises  the  great  square  ABCD. 

From  either  of  the  preceding  modes  of  examining  this 
subject,  we  draw  the  following 

Rule.  \^  Separate  the  given  number  into  periods^  of 
two  fici-ures  each,  beginning  at  the  iiniCs  place. 

2.  Find  the  greatest  square  contained  in  the  left-hand 
period,  and  set  its  root  on  the  right  of  the  given  num- 
ber; sulHract  said  square  from  the  left-hand  period,  and 
to  the  remainder  bring  down  the  next  period  for  a  di- 
vidual. 

3.  Do2(ble  the  root  for  a  divisor,  and  try  how  often 
this  divisor  {with  the  figure  used  in  the  trial  thereto  an- 
nexed) is  contained  in  the  dividual:  set  the  number  of 
times  in  the  root ;  then,  multiph/  and  subtract  as  in  di- 
vision, and  bring  down  the  next  period  to  the  remain- 
der for  a  new  dividual. 

4.  Double  the  ascertained  root  for  a  new  divisor,  and 
proceed  as  before,  till  all  the  periods  are  brought  down. 

[Note.  When  all  the  periods  are  brought  down,  if 
there  be  a  remainder,  annex  ciphers  to  it  for  decimals, 
and  proceed  till  the  root  is  obtained  to  a  sufficient  de- 
gree of  exactness. 

Observe,  that  the  decimal  periods  are  to  be  pointed  off 
from  the  decimal  point  toward  the  right  hand  ;  and  that 
there  must  be  as  many  whole  number  figures  in  the  root, 
as  there  are  periods  of  whole  numbers,  and  as  many  de- 
cimal figures  as  there  are  periods  of  decimals.] 

EXAMPLES. 

1.  What  is  the  square  root  of  729?  Jlns.  27. 

2.  What  is  the  square  root  of  106929  ?  Ans.  327. 

3.  What  is  the  square  root  of  429025  ?  Ans.  655. 

4.  What  is  the  square  root  of  444889  ?  Ans.  667. 

5.  What  is  tlie  square  root  of  776161  ?  Ans.  881. 

6.  What  is  the  square  root  of  994009  ?  Ans.  997. 

7.  What  is  the  square  root  of  29855296  ? 

Ans.  5464. 

8.  What  is  the  square  root  of  141225,64  ? 

Ans.  375,8. 

u"2 


234  ARITHMETIC. 


9.  What  is  the  square  root  of  43046271  ? 

.fins.  6561. 

10.  What  is  the  square  root  of  904? 

^ns.  30,066  + 

11.  What  is  the  square  root  of  670? 

MS.  1^,884  -}- 

12.  What  is  the  square  root  of  20? 

^ns.  4,472  + 

13.  What  is  the  square  root  of  2  ? 

^ns.  1,4142  + 

14.  Find  the  square  root  of  3.  .dns.  1.732  -f- 

15.  Find  the  square  root  of  5.  ^ns.  2,236  -j- 

16.  Find  the  square  root  of  10.  j9ns.  3,1822  + 
N.  B.   When  the  power  is  a  decimal,  commence  to 

point  off  periods  from  the  decimal  point,  and  operate  as 
in  whole  numbers. 

EXAMPLES. 

1.  What  is  the  square  root  of  ,00008836? 

^7is.  ,0094. 

2.  Find  the  square  root  of  ,00529.  Jins.  ,023. 

3.  Find  the  square  root  of  ,002916.  Jiyis.  ,054. 
Observe,  the  roots  are  greater  than  the  powers,  as  they 

should  be  in  fractions. 

(Art.  129.)  Mixed  numbers  may  be  square,  when  re- 
duced to  the  fractional  form  (improper  fractions)  ;  but  the 
square  form  will  commonly  be  indicated  by  the  denomi- 
nator of  the  fraction  being  a  square  number.  In  such 
cases,  reduce  the  mixed  numbers  to  improper  fractions, 
and  extract  the  roots  of  numerator  and  denominator. 

EXAMPLES. 

1.  What  is  the  square  root  of  37^f  ? 

2.  What  is  the  square  root  of  272  \-  ? 

3.  Find  the  square  root  of  27yg. 

4.  Find  the  square  root  of  9i|. 

5.  Find  the  square  root  of  20'^.  .^ns.  4i. 
Fractions  may  be  square  in  value,  but  not  expressed 

by  square  numbers;  but  can  be  reduced  to  such,  and  af- 


Jlns. 

61. 

Mns. 

16L. 

Ans. 

5|. 

Arts. 

31. 

SQUARE    ROOT.  235 


terwards  the  square  root  extracted.     The  following  ex- 
amples will  illustrate  : 

6.  What  is  the  square  root  of  -,~  ?  ^dns.  i. 

7.  What  is  the  square  root  of  f^^  ?  Jinn,  f . 

8.  Find  the  square  root  of  =-|^ .  Ans.  ^. 

9.  What  is  the  square  root  of  ^^5.|  ?  ^ns.  |. 

10.  What  is  the  square  root  of  f  l||  ?  Jins,  |. 

11.  Find  the  square  root  of  ||-|J.  Ans.  |. 

When  fractions  or  mixed  numbers  are  surds,  reduce 
the  fractions  to  decimals,  and  proceed  as  in  whole  num- 
bers. 

EXAMPLES. 

1.  What  is  the  square  root  of  7/^  ?  Ans.  2,7961 -f 

2.  Find  the  square  root  of  17|.  Ans.  4,168-}- 

3.  Find  the  square  root  of  ?.  Ans.  ,86602  -[- 

4.  Find  the  square  root  of  ~\.  Ans.  ,9574-}- 

5.  Find  the  square  root  of  85{-i.  Ans.  9,27-f- 

APPLICATION. 

1.  A  certain  square  pavement  contains  20736  square 
stones,  all  of  the  same  size  ;  what  number  is  contained  in 
one  of  its  sides?  Ans.  144. 

2.  If  484  trees  be  planted  at  an  equal  distance  from 
each  other,  so  as  to  form  a  square  orchard,  how  many- 
will  be  in  a  row  each  way  ?  Ans.  22. 

3.  A  certain  number  of  men  gave  30  shillings  1  pence 
for  a  charitable  purpose  ;  each  man  gave  as  many  pence 
as  there  were  men  :  how  many  men  were  there  ? 

A7ts.  19. 

(Art.  130.)  To  be  scientific  and  skillful  in  the  appli- 
cation of  square  root,  it  is  necessary  to  attend  to  the  fol- 
lowing properties  of  square  numbers. 

I.  A  square  number,  multiplied  by  a  square  number, 
will  produce  a  square  number. 

II.  A  square  number,  divided  by  a  square  number  will 
give  a  square. 

III.  If  the  square  root  of  a  number  is  a  composite 


236  ARITHMETIC. 


number,  the  square  itself  may  be  divided  into  integer 
square  factors  ;  but  if  the  root  is  a  prij7ie  number,  the 
square  cannot  be  separated  into  square  factors,  ivitkout 
fractions. 

IV.  When  two  or  more  factors  are  to  be  multiplied  to- 
gether, we  may  often  abbreviate  the  operation,  by  ob- 
serving whether  they  are,  or  whether  we  can  change  them 
into,  square  factors.  If  so,  take  the  square  roots  of  the 
factors,  and  multiply  them  together  for  the  result. 

EXAMPLES. 

1.  A  section  of  land  in  the  western  states,  is  a  square 
consisting  of  640  acres  ;  what  is  the  length  in  rods,  of 
one  of  its  sides?  ^^ns.  320. 

Very. many  of  our  teachers  7,oouJd  actually  reduce  the 
acres  to  square  rods,  by  multiplying  by  160,  and  extract 
the  square  root  of  the  product ;  but  this  would  show  too 
little  attention  to  numbers.  Remove  one  of  the  ciphers 
from  one  number  to  the  other,  and  we  have  64  to  be  mul- 
tiplied by  1600,  both  square  numbers,  whose  roots  are  8 
and  40— product  320,  the  answer. 

2.  A  man  has  a  farm  in  exactly  a  square  form,  con- 
taining 50|  acres  ;  how  many  rods  on  a  side  ? 

Ans.  90  rods. 
Solution,  50|  =  ^|-^    Then  ^f^  x  160=405X20=81 
XIOO. 

3.  A  certain  triangular  field  contains  10  acres  of  land  ; 
what  is  the  length  of  the  sides  of  a  square  containing  the 
same  number  of  acres  ?  Ans.  40  rods. 

4.  What  must  be  the  side  of  a  square  field,  that  shall 
contain  an  area  equal  to  another  field  of  rectangular 
shape,  the  two  adjacent  sides  of  which  are  18  by  72 
rods  ?  Ans.  36  rods. 

18  by  72  is  the  same  as  the  half  of  18  by  the  double 
of  72,  or  9  by  144,  square  numbers,  roots  3X  12=36, 
the  answer. 

5.  A  has  two  fields,  one  containing  10  acres,  and  the 
other  12  V  ;  what  will  be  the  length  of  the  side  of  a  field 
containing  as  many  acres  as  both  of  them? 

Ans.  60  rods. 


SQUARE    ROOT.  237 


6.  It  is  required  to  lay  out  120  acres  of  land  in  the 
form  of  a  right-angled  parallelogram,  so  that  the  length 
shall  be  3  times  the  breadth  ;  required  the  sides. 

.^iis.  240  feet  length,  80  feet  i^readth. 

7.  Required  the  dimensions  of  a  square,  the  area  of 
which  shall  be  equal  to  the  area  of  a  parallelogram  18 
feet  in  length,  and  8  in  breadth.     Jlns,  12  feet  square. 

(Art.  131.)  A  mean  proportional  between  two  num- 
bers, is  a  number  requisite  to  form  the  means  of  a  geo- 
metrical proportion  between  those  numbers.     Thus, 

3  :  6  ::  6  :   12 
The  two  means  in  this  proportion  are  expressed  by  the 
some  nwnber.    That  number  is,  therefore,  called  a  mean 
proportional,  and  its  square  is  equal  to  the  product  of  the 
given  extremes  (Art.  72). 

Hence,  to  find  a  mean  proportional  between  two  num- 
bers, we  may  have  this 

Rule.  Mulliply  the  numbers  together,  arid  take  the 
square  root  of  the  product. 

Observation.  When  the  given  numbers  are  square 
numbers,  take  their  roots  at  once,  and  multiply  them  to- 
gether ;  or,  if  convenient,  change  them  into  equivalent 
square  factors,  &;c.,  as  in  Art.  130. 

EXAMPLES. 

1.  Find  the  mean  proportional  between  4  and  25. 

Ans.  10. 

2.  Find  the  mean  proportional  between  2  and  12,5. 

Ans.  5. 

3.  Find  the  mean  proportional  between  ^  and  4. 

Jins.  |. 

4.  Find  the  proportional  between  24  and  96. 

Jlns.  48. 
As  96=4X24;  the  product  of  the   two  numbers  is 
equal  to  4X24X24=4X  24^  root,  is  2X24=48  Ans, 

5.  Find  the  mean  proportional  between  18  and  24. 

Am.  1273. 

6.  Find  the  mean  proportional  between  18  and  32. 

Arts,  24. 


238  ARITHMETIC. 


7.  Find  the  mean  proportional  between  ,25  and  1. 

Ans.  ,5. 

8.  AVhat  is  the  mean  proportional  between  3  and  6  1 

Arts.  SJ2. 

9.  What  is  the  mean  proportional  between  4  and  40  ? 

Ans.  4^10. 

These  surd  answers  may  be  taken  more  definitely  by 
extracting  the  root  in  decimals,  or  by  taking  the  roots  al- 
ready extracted  in  Art.  128. 

(Art.  132.)  A  right-angled  triangle,  is  a  triangle  hav- 
ing one  right  angle,  and  the  other  two  angles  acute.  The 
side  opposite  the  right  angle  is  called  the  hypotenuse. 

C 

The  adjoining  figure  represents  a  right- 
angled  triangle — the  right  angle  at  B.  It 
is  proved,  in  geometry,  that  the  sum  of 
the  squares  of  the  two  sides  containing  the 
right  angle,  is  equal  to  the  square  of  the 
side  opposite  the  right  angle.  Therefore 
the  difference  of  the  squares  between  the     B  A 

hypotenuse  and  one  of  the  sides,  is  equal  to  the  square 
of  the  other  side. 

It  is  also  proved,  in  geometry,  that  triangles  of  the 
same  shape  have  their  like  sides  proportional. 

This  last  theorem  enables  us  to  abbreviate  and  reduce 
large  right-angled  triangles  to  smaller  ones,  and  the  re- 
verse. 

PROBLEMS    ON    THE    RIGHT-ANGLED    TRIANGLE. 

1.  The  top  of  a  castle  is  45  yards  high,  and  is  sur- 
rounded with  a  ditch  60  yards  wide ;  required  the  length 
of  a  ladder  that  will  reach  from  the  outside  of  the  ditch 
to  the  top  of  the  castle.  Ans.  75  yards. 

This  is  almost  invariably  done  by  squaring  45  and  60, 
adding  them  together,  and  extracting  the  square  root ;  but 
so  much  labor  is  never  necessary  ivhen  the  numbers  have 
a  common  divisor,  or  when  the  side  sought  is  expressed 
by  a  composite  number. 

Take  45  and  60  ;  both  may  be  divided  by  15,  and  they 
will  be  reduced  to  3  and  4.     Square  these,  9-|- 16=25. 


SQUARE    ROOT.  239 


The  square  root  of  25  is  5,  wliich,  multiplied  by  15,  gives 
75,  the  Miiswer. 

2.  Two  brothers  left  their  father's  house  and  went,  one 
64  miles  due  west,  the  other  48  miles  due  north,  and 
purchased  farms  ;  how  far  are  they  from  each  other  ? 

^^ns.  80  miles. 

Divide  by  ...  16)64  48 


4     3     164-9=25,  5X16=80. 

3.  The  hypotenuse  of  a  riglu-angled  triangle  is  520 
feet,  the  base  312  feet;  what  is  the  perpendicular  ? 

Ans.  416. 
Divide  by.  .  .   52)520  312 


2)10       6 

5       3     25—9  =  16,  root  4. 
Multiply  by 104 

Answer 416 

4.  Required  the  height  of  a  May-pole,  whose  top  being 
broken  off,  struck  the  ground  at  the  distance  of  15  feet 
from  the  foot,  and  measured  39  feet.         Ans.  75  feet. 

3.  A  hawk,  perched  on  the  top  of  a  perpendicular  tree, 
77  feet  high,  was  brought  down  by  a  sportsman,  standing 
off  14  rods,  on  a  level  with  its  base;  what  distance,  in 
yards,  did  he  shoot?  Ans,  81,15+yards. 

If  this  problem  is  worked  with  skill,  it  will  be  requi- 
site to  extract  the  root  of  10  only. 

6.  If  the  diagonal  of  a  rectangular  field  is  40  rods,  and 
one  of  the  sides  32,  what  is  the  other?  Ans.  24. 

When  the  hypotenuse  and  one  side  is  given,  in  place 
of  squaring  them,  and  taking  the  difference,  &c.,  we  may 
take  the  square  root  of  the  product  of  their  sum  and 
difference,  for  the  answer,  as  in  the  following  example : 

7.  A  line  27  yards  long,  will  exactly  reach  from  the 
top  of  a  fort  to  the  opposite  bank  of  a  river,  which  is 
known  to  be  23  yards  broad;  what  is  the  height  of  the 
fort?  Ans.  14,142-+-yards. 


340  ARITHMETIC. 


Sum,  50;  diff.  4;  ^50X4  =  V200  =  14,142  +  ^n.s. 

8.  Suppose  a  ladder  40  feet  long  be  so  planted  as  to 
reach  a  window  33  feet  from  the  ground,  on  one  side  of 
the  street,  and  witliout  moving  it  at  the  foot,  will  reach  a 
window  on  the  other  side  21  feet  high,  what  is  the 
breadth  of  the  street  ?  ^iis.  56,64-4-feet. 

9.  A  ladder  40  feet  long  stands  close  against  the  side 
of  a  building;  it  is  drawn  out  at  the  bottom  sufliciendy 
to  lower  the  top  2  feet;  how  many  feet  is  it  drawn  out 
at  bottom  ?  ^^^^  ^156= 12,49  feet,  nearly. 

10.  A  certain  room  is  28  feet  long,  22  feet  wide,  and 
12  feet  high  ;  what  is  the  diagonal  distance  from  the  low- 
er to  the  opposite  upper  corner  of  the  room  ? 

.^7is,  37,57+feet. 

11.  Wishing  to  know  the  distance  from  a  station  near 
the  side  of  a  river  to  a  tree  below,  but  being  prevented 
by  a  curve  in  the  stream  from  measuring  it,  I  run  out  at 
right  angles  to  the  line  of  direction  15  rods,  and  then  30 
rods  in  a  right  line  to  the  tree;  what  was  the  distance 
required  ?  Jins.  26  rods,  nearly. 

12.  Suppose  the  width  of  a  barn  to  be  36  feet,  of  what 
length  must  the  rafters  be,  if  the  elevation  of  the  ridge 
above  the  level  of  their  foot  be  14  feet. 

Ms.  22  feet  9|  in. 

Questions. 

What  is  evolution  ? 

To  extract  the  square  root,  why  do  you  point  off  pe- 
riods of  2  figures  each? 

In  obtaining  divisors,  why  do  you  double  the  root  al- 
ready found  ? 

Has  every  number  an  exact  root  ? 

What  is  a  right-angled  triangle  ? 

What  relation  is  there  between  the  square  of  the  hy- 
potenuse, and  the  squares  of  the  two  sides  ? 

If  the  two  sides  are  given,  how  do  we  find  the  hypo- 
tenuse ? 

If  the  hypotenuse  and  base  are  given,  how  is  the  per- 
pendicular found  ? 


CUBE    ROOT.  241 


If  the  hypotenuse  and  perpendicular  are  known,  how 
do  we  find  the  base  ? 

Can  you  ever  reduce  a  right-angled  triangle  to  smaller 
dimensions  ? 

Can  you  reduce  a  right-angled  triangle,  v/lien  the  num- 
bers expressing  the  sides  are  prime  numbers? 

How  do  you  return  to  the  original  triangle  ? 

TO    EXTRACT    THE    CUBE    ROOT. 

(Art.  133.)  The  cube  of  a  number  is  the  product  of 
that  number,  multiplied  into  its  square.  A  cube  is  also 
a  geometrical  solid,  of  six  equal  square  surfaces.  From 
each  of  these  definitions,  or  modes  of  considering  a  cube, 
we  can  draw  the  same  rule  for  extracting:  the  root;  one, 
by  retracing  the  multiplication  of  the  different  figures  of 
the  root,  to  form  the  cube:  the  other,  by  constructing  and 
increasing  geometrical  solids:  we  shall  orive  the  analysis 
of  both. 

In  Art.  126,  we  have  sufficiently  explained  that  the 
cube  of  a  number  may  contain  three  times  the  number  of 
places  of  figures  that  are  in  the  root ;  and,  therefore,  we 
must  divide  a  cube  off  into  periods  of  three  figures  each, 
to  fiiid  the  number  of  figures  in  the  root;  but  t!iis  we 
deem  of  sufiicient  importance,  to  take  another  view  of 
the  subject.  If  we  take  any  figure  (say  8),  and  involve 
it  as  units,  tens,  hundreds,  and  so  on:  that  is,  as  8,  80, 
800,  and  so  on;  we  shall  readily  discover  the  Inw  v.'hich 
governs  tlie  posiiion  of  the  successive  products,  8  X  8X  8 
=  512.  As  a  unit,  its  cube  terminates  with  the  unit's 
place,  and  influences  hut  two  other  places,  the  tens  and 
liundreds.  As  a  ten,  its  proper  product  is  the  same,  but 
is  removed  three  places  toward  tlie  left;  thus,  80X80X 
80  =  512000.  As  a  hundred,  its  product,  a  cube,  \\\\\ 
fall  three  places  higher,  and  so  on — clearly  indicating 
three  places  in  a  cube  to  one  in  the  root. 

(Art.  134.)  To  trace  back  a  cube,  let  us  involve  a 
number,  consisting  of  two  figures,  to  its  cube,  keeping 
the  numbers  separate  by  signs,  and  observing  that  a  num- 
ber multiplied  by  itself,  gives  its  square,  which  may  be 
indicated  by  the  exponent  2,  the  cube  by  the  exponent  3. 


242  ARITHMETIC. 

Cube  46.  Express  it  thus,  .  .  .  •  40-r6 

40+6 

40X6-1-62 

40^-1-       40X6 

402-j-  2  X  40  X  6+63.=square. 

40-1-6 

40^X6+2X40X62-1-65 
40^+2X40^X6+       40X6^ 

40^+3  X  40^  X  6+3  X  40  X  62+6-=cube. 
The  whole  cube  then  consists  of — 

1.  The  cube  of  the  tens 40^         =64000 

2.  Three  times  the  square  of  the 

tens  into  the  units, 3  X  40""  X  6  =28800 

3.  Three  times  the  tens  into  the 

square  of  the  units, 3X40  X  6'=  4320 

4th,  and  last.  The  cube  of  the 

units, 6^=     216 

Entire  cube, =97336 

To  retrace  this  operation — to  extract  the  cube  root  of 
97-336 — we  must  divide  it  off  into  periods,  of  three  fig- 
ures each,  commencinfT  at  the  units  ;  this  gives  97  for  the 
superior  period,  in  which  the  cube  of  the  tens  must  ex- 
ist. Take  die  greatest  possible  cube  in  97,  which  is  64, 
(whose  root  is  4,)  and  subtract  it  from  the  whole  cube ; 
thus, 

97*336(4 

64 

33336= 1st  dividual. 

Let  us  call  to  mind  the  fact,  that  we  know  the  root  of 
this  cube  already,  and  that  its  next  superior  constituent 
part  is  3.4016."  Observe,  that  402=4-X  100,  and  this, 
multiplied  by  3,  is  4^X300;  that  is,  the  square  of  the 
root  already  found,  multiplied  by  300.  In  retracing,  the 
6  is  not  yet  discovered  ;  but  42X  300=4800,  forms  a  par- 


CUBE    ROOT.  2'13 


lial  divisor  for  tlie  dividual.  Tlie  number  of  times  this 
will  go  in  the  dividual,  is  an  index  for  the  next  figure. 
Wlieii  the  divisor  is  small  in  relation  to  the  dividual,  the 
next  ligurc  may  be  one  or  two  uniis  less  than  the  quo- 
tient, as  the  dividual  must  contain  three  terms.     Now, 

97336(46 

64 

Partial  divisor,  4800  )33336=dividual. 

4800X0  28800  =  lstterm. 

4  X  6- X  30  =3.40.62=       4320 =2d  term. 

6X6X6  =        216  =  3d  and  last  term. 


S3336=the  dividual. 
This  operation,  put  in  words,  gives  the  following 

RuLR.  Separate  the  given  number  into  periods  of 
three  figures  each,  beginning  at  the  iinii's  place.  Find 
the  greatest  cube  in  the  left-hand  period,  and  set  its 
root  in  the  quotient;  subtract  said  cube  from  the  pe- 
riods, and  to  the  remainder  bring  doivn  tlie  next  pe- 
riod for  a  dividual.  Square  the  root  cdready  found, 
and  multiply  it  by  300, /or  a  partial  or  imperfect  divi- 
sor. See  how  often  the  partial  divisor  is  contained  in 
the  dividucd,  and  take  the  result  for  the  next  figure  of 
the  roof. 

[Note.  As  the  divisor  is  partial,  7?o?/^//,  this  quotient, 
ill  some  cases,  will  be  too  large  for  the  next  figure  of  th.e 
root:  there  is  no  certain  way  of  deciding  except  by  trial.] 

Multiply  the  particd  divisor  by  the  last  foitnd  figure 
of  the  root ;  also  multiply  the  square  of  the  last  found 
figure  by  the  former  part  of  the  root,  and  that  product 
by  30  ;  ab-o  cube  the  last  found  figure.  .,^dd  these  three 
products  together,  and  subtract  their  amount  fr era  tJie 
dividual. 

To  the  remainder,  bring  dcicn  the  next  period,  and 
proceed  as  before,  until  the  periods  are  all  brought  down. 

TVhcn  (I  remainder  occurs,  annex  periods  of  ciphers 
to  obtain  decimals,  which  may  be  carried  to  any  desired 
number. 


244 


ARITHMETIC. 


N.  B.  Tlie  cube  root  of  a  vulgar  fraction  is  found  by 
reducing  it  to  its  lowest  terms,  and  extracting  the  root  of 
tlie  numerator  for  a  numerator,  and  of  the  denominator 
for  a  denominator.  If  it  be  a  surd,  extract  the  root  of  its 
equivalent  decimal  to  the  desired  degree  of  exactness. 

(Art.  135.)  We  now  illustrate  by  means  of  geometri- 
cal solids.     What  is  the  cube  root  of  15625  ? 


Operation, 


2X2X2=  8 


15,625  1 25 


7625 


In  this  number  there 
are  2  periods  ;  of  course 
there  will  be  2  figures 
in  the  root.  

''The  greatest  cube  2X2X300  =  1200  |  7625 

in  the  lefi-hand  period  

(15),  is  8,  the  root  of  6000 

i^,'/u*cA  fs  2  ;"  therefore,         5X5X2X30=   1500 
2  is  the  hrst  figure  of  5X5X5=      125 

the    root ;    and,   as  we  

shall  have  another  fig- 
ure in  the  root,  the  2 
stands  for  lens,  or  20. 
But  the  cube  root  is  the 
length  of  one  of  the 
sides  of  the  cube,  whose 
length,  breadth,  and 
thickness,  are  equal : 
then  the  cube  whose 
root  is  20,  contains  20 
X20X20=8000. 

"  Subtract  the  cube 
thus  found  (8)  from 
said  periods  and  to  the 
remainder  bring  doicn  the  next  period'''  or,  subtract 
the  8000  from  the  whole  given  number  (15625),  and 
7625  will  remain.  Thus  8000  feet  are  disposed  of  in 
the  cube,  Fig.  1,  20  feet  long,  20  feet  wide,  and  20  feet 
high. 

The  cube  is  to  be  enlarged  by  the  addition  of  7625 
feet  which  remain.  In  doing  this,  the  figure  must  be  en- 
larged on  three  sides,  to  make  it  longer,  and  ivider,  and 
higher,  to  maintain  the  complete  cubic  form. 


CUBE    ROOT. 


245 

The  next  step  is,  to  find  a  divisor;  and  tliis  mnst  be 
the  number  of  square  feet  contained  in  the  three  sides  to 
which  the  addition  must  be  niadc.  Hence  we  "  muUlphi 
the  square  of  the,  quotient  /i<ri(re  by  300."  TJiat  is, 
2X2X300=1200:  or,  20X26  X  3  =  1200  feet,  which  is 
the  superficial  content  of  the  tliree  sides,  «,  b,  and  c. 

This  ''divisor  (1200)  is  con- 
tained inthc.dividiiaP'  (7625) 
5  times:  tlien  5  is  the  second 
quotient  fitrure;  that  is,  the  ad- 
dition to  each  of  the  three  sides 
is  5  feet  thick;  if  1200  feet  co- 
ver the  three  sides  1  foot  thick, 
5  feet  thick  will  require  5  times 
as  many;  that  is,  1200X5  = 
6000. 

But  when  tlie  additions  are 
made  to  three  squares,  there 
will  be  a  deficiency  along  the 
whole  length  of  the  sides  of 
the  squares  between  the  addi- 
tions, which  must  be  supplied 
before  the  cube  will  be  com- 
plete. These  deficiencies  will 
be  three,  as  may  be  seen  at 
NNN  in  Fig.  2;  therefore  it 
is  that  we  "■midtiply  the  square 
of  the  last  figure  by  tlie  prece- 
ding figure,  andtnj  30,"  (ihat 
is,  5  X  5  X  2  X  30,)  or,  5  X  5  X  5 
X20X3  =  1500,  which  is  the 
quantity  required  to  supply  the 
three  deficiencies. 

Figure  3  represents  the  sol- 
id, with  these  deficiencies  sup- 
plied, and  discovers  another  de- 
ficiency, where  they  approach 
each  other  at  000. 

Lastly,  ''cube  the  last  fg- 
ure:^'  this  is  done  to  fill  the  de- 


Hi 


x2 


246 


ARITHMETIC. 


Proof. 

20X20X20=  8000 

20X20X3X5=  6000 

5X5X20x3=  1500 

5X5X5=  125 

25X25X25=  15625 
quently,  the  cube,  of 


ficiency  left  at  the  corner, 
in  filling  up  the  other  defi- 
ciencies. This  corner  is 
limited  by  the  three  por- 
tions applied  to  fill  the  for- 
mer vacancies,  which  were 
5   feet  in   breadth :    conse- 


5  will  be  the  solid  contents  of  the 
corner.  Fig.  4  represents  this  deficiency  [eee)  supplied, 
and  the  cube  is  complete. 

Observe,  that  20  X  20  X  3  X  5  =2^  X  300  X  5  ;  and  thus, 
by  a  Utile  change  of  form,  the  numerical  and  geometrical 
analysis  agree,  and  will  give  one  and  the  same  rule. 


EXAMPLES. 

1.  AVhat  is  the  cube  root  of  99252847? 
99'252'847(463 
4X4X4=64 


4X4X300=4800 

Div.  4800X6= 

6X6X4X30  = 

6X6X6= 


35252 

28800 

4320 

216 


.^ns.  463. 


463 
463 


Subtrahend,  33336 


46X46X300=634800  1  1916847 


1389 

2778 
1852 

214369 
463 


Div.  634800X3=1904400                    613107 

3X3X46X30=      12420                 1286214 

3X3X3=                      27                857476 

Subtrahend,          1916847    Proo/,  99252847 

2. 

What  is  the  cube  root  of  84604.519?  Ans.  439. 

3. 

What  is  the  cube  root  of  259694072  ? 

^      ■     Ans.  638. 

4. 

What  is  the  cube  root  of  32461759  ? 

A71S.  319. 

5. 

What  is  the  cube  root  of  27054036008  ? 

i 

Ans.  3002. 

CUBE   ROOT.  247 


6.  What  is  the  cube  root  of  22069810125? 

Ans.  2805. 

7.  What  is  the  cube  root  of  15926,9725  ? 

Ans.  25,16+ 
Wheii  decimals  occur,  point  the  periods  both  ways, 
beginning  at  the  decimal  point :  and  if  the  last  period  of 
the  decimal  be  not  complete,  arr.ex  one  or  more  ciphers. 

A  mixed  number  may  be  reduced  to  an  improper  frac- 
tion, or  a  decimal,  and  then  the  root  extracted. 

8.  What  is  the  approximate  cube  root  of  ,000684134  ? 

Ans.  ,088-f 

9.  Find  the  approximate  cube  root  of  ,008649. 

Ans.  ,2052-1- 

10.  What  is  the  cube  root  of  /jVo?  ^^»«-  f 

11.  What  is  the  cube  root  of  |f^?  Ans.  |. 

12.  What  is  the  cube  root  of  W^l  A71S.  f. 

13.  What  is  the  cube  root  of  12^^  1  Ans.  2L. 

14.  What  is  the  cube  root  of  31 3' 4V-  ^^'^^'  ^f- 

APPLICATION. 

1.  There  is  a  cistern,  or  vat,  of  a  cubical  lorm,  which 
contains  1331  cubical  feet;  what  are  the  lenglh,  breadth 
and  depth  of  it?  Ans.  Each  11  feet. 

2.  A  certain  stone  of  a  cubical  form  contains  474552 
solid  inches  ;  what  is  the  superticial  content  of  one  of  its 
sides,  expressed  in  feet?  Ans.  421  sq.  feet. 

3.  The  pedestal  of  a  monument  was  a  square  block 
of  gi-anite,  coniaining  373248  cubic  inches;  what  was 
the  length  of  one  of  its  sides  ?  Ans.  G  feet. 

4.  The  statute  bushel  contains  2150,4  cubic  incjies  ; 
what  is  the  side  of  a  cubic  box  that  will  contain  50  bush- 
els ?  Ans.  47,5  inches. 

5.  A  man  wishes  to  make  a  bin  to  contain  125  bushels, 
of  equal  depth  and  width,  and  double  in  length ;  what 
shall  be  its  dimensions  ? 

Ans.  Width  and  depth,  5 1,22+ inches. 
In   lenglh,  ....   102,44  inches. 

(Art.  136.)  It  is  proved  in  solid  geometry,  that  all  so- 
lid bodies  of  similar  shape,  are  to  one  another  as  the 
cubes  of  their  like  dimensions. 


248  ARITHMETIC. 


N.  B.  We  need  not  take  their  real,  if  we  can  find 
their  more  simple,  relalive  dimensions. 

On  this  principle,  solve  the  following  problems  : 

1.  Mercury  is  about  2000  miles  in  diameter,  and  the 
earth  about  8000  ;  what  is  their  relative  magnitudes  ? 

.r2ns.  As  1  to  64. 

2.  Mars  is  about  4000  miles  in  diameter,  the  earth 
8000  ;  what  is  their  relative  magnitudes  ? 

.^728.  As  1  to  8. 

3.  The  diameter  of  the  earth,  to  that  of  the  sun,  is 
nearly  as  1  to  111!,;  what  is  their  relative  magnitudes, 
or  bulks?  "        .^??5.  As  1  to  13fGUi5,  nearly. 

4.  If  a  ball,  6  inches  in  diameter,  weigh  32  pounds, 
what  will  be  the  weight  of  a  ball,  of  the  same  metal, 
whose  diameter  is  3  inches  ?  ^^tis.  4  pounds, 

6     :     3     ::     2     :     I 
8     :      1      :  :  32     :     4 

5.  Suppose  an  iron  ball,  of  4  inches  in  diameter,  to 
weigh  9  pounds  ;  required  the  weight  of  a  spherical  shell 
of  9  inches  in  diameter,  and  1  inch  thick. 

./^ns.  54  lbs.  4  oz. 

6.  If  a  cable,  12  inches  about,  require  an  anchor  of 
18  hundred  weight,  of  what  weight  must  an  anchor  be 
for  a  15  inch  cable  ?  .^ns.  35  cut.  15  lbs. 

7.  Suppose  the  diameter  of  the  earth  to  be  3952  miles, 
and  a  portion  of  it  is  water  sufficient  to  cover  the  whole 
body  to  the  average  depth  of  9\  miles  ;  what  proportion  of 
the  earth's  bulk  would  then  be  water  ?  1 

»^ns.     n. 

69 

8.  If  a  ball  of  lead,  2  inches  in  diameter,  Aveighs  2 
pounds,  what  will  be  the  weight  of  another,  10  incb^es  in 
diameter  ?  .^ns.  250  Ids. 

9.  In  2  cubic  feet,  how  many  cubes  of  6  inches  ? 

»^ns.  16. 


SUPPLEMENT    TO    SqUARK    AND    CUBE    ROOTS. 


249 


SUPPLEMENT  TO  SQUARE  AND  CUBE 
ROOTS. 

RECAPITULATION. 

We  again  call  to  mind  the  following  properties  of  num- 
bers. Their  importance  cannot  be  exaggerated,  if  we  wish 
to  insure  skill,  or  even  sound  knowledge,  on  this  subject. 

I.  A  square  number,  multiplied  by  a  square  number, 
the  product  will  be  a  square  number. 

II.  A  square  number,  divided  by  a  square  number,  the 
quotient  is  a  square. 

III.  A  cube  number,  multiplied  by  a  cube,  the  product 
is  a  cube. 

IV.  A  cube  number,  divided  by  a  cube,  the  quotient 
will  be  a  cube. 

V.  If  any  root  is  a  composite  number,  its  power  (the 
square  or  cube,  as  the  case  may  be)  may  be  separated 
into  square  or  cube  factors;  but,  if  the  root  is  ^ prime 
number,  the  power  cannot  be  so  separated. 

VI.  If  the  unit  figure  of  a  square  number  is  5,  we  may 
multiply  by  the  square  number  4,  and  we  shall  have  an- 
otlier  square,  whose  ?//n7  period  will  be  ciphers. 

VII.  If  the  unit  figure  of  a  cube  is  5,  we  may  multiply 
by  the  cube  number  8,  and  produce  another  cube,  whose 
unit  period  will  be  ciphers. 

N.  B.  If  a  supposed  cube,  whose  unit  figure  is  5,  be 
multiplied  by  8,  and  the  product  does  not  give  three  ci- 
phers on  the  right,  the  number  is  not  a  cube. 

(Art.  137.)  V/e  again  present  the  following  table,  for 
the  pupil  to  compare  the  natural  numbers  with  the  unit 
figure  of  their  squares  and  cubes,  that  he  may  be  able  to 
extract  roots  by  inspection. 


Numbers, 

I 

2 

3   4 

1 

5 

6 

7 

8 

9 

10 

Squares. 

1 

4 

9 

IG 

25 

36 

49 

64 

81 

100 

Cubes. 

1 

8  27 

64 

125 

216 

343 

512 

729 

1000 

250  ARITHMETIC. 


EXERCISES    FOR    PRACTICE. 

1.  What  is  the  square  root  of  G25?  ^%is.  25. 
If  the  root  is  an  integer  number^  we  may  know,  by 

the  inspection  of  the  table,  that  it  must  be  25,  as  the 
greatest  square  in  6  is  2,  and  5  is  the  only  figure  whose 
square  is  5  in  its  unit  place. 

Again,  take 625 

Multiply  by 4     4  being  a  square. 

2500 

The  square  root  of  this  product  is  obviously  50  ;  but, 
this  must  be  divided  by  2,  the  square  root  of  4,  which 
gives  25,  the  root. 

2.  What  is  the  square  root  of  5561  ?  Ans.  81. 

As  the  unit  Jigiire,  in  this  example,  is  1,  and  in  the 
line  of  squares  in  the  table,  we  find  1  only  at  1  and  81, 
we  will,  therefore,  divide  6561  by  81,  and  we  find  the 
quotient  81 ;  81  is,  therefore,  the  square  root. 

3.  What  is  the  square  root  of  106729  ?     Ans.  327. 

As  the  unit  figure,  in  th^s  example,  is  9,  if  the  number 
is  a  square,  it  si  divisible  by  either  9,  or  49.  After  di- 
viding by  9,  we  have  11881  for  the  other  factor,  a  prime 
number;  therefore  its  root  is  a  prime  number=lG9.  109, 
multiplied  by  3,  the  root  of  9,  gives  327  for  the  answer. 

4.  What  is  the  mot  of  451584  ?  Ans.  672. 
This  number  is  obviously  divisible  by  the  square  num- 
ber 4:  thus,  4)451584 

Divide  again  by  4, 4)112896 

And  again, 4)28224 

And  again, 4^7056 

And  again,  as  long  as  the  division  is  obvious,  4)1764 

441 
Now,  the  square  rcot  of  the  last  quotient  is  21,  and 
the  square  roots  of  the  divisors  are  2.  2,  2,  2,  2;  the  con- 
tinued product  of  all  these  roots  is  672,  the  ansv/er. 


ABBREVIATIONS    IX    CUBE    ROOT.  251 

5.  Extract  the  square  root  of  21)25.  Ans.  Ad. 
1st.    Divide  by  the  square  number,  25,  and  we  tind 

tlie  two  factors,  25X81,  as  equivalent  to  the  given  num- 
ber.    Roots  of  these  factors,  5X9  =  45,  the  answer. 

2d.  Divide  by  tlie  square  number,  100,  and  we  have 
20^X100  =  ^X100,^X10  =  45,  as  before. 

3d.  Take .  .  .  2025 
Multiply  by  .  .  .        4 

8100  Root  90,  divide  by  2,  because 
we  multiplied  the  power  by  4,  and  we  have  45  for  the 
result. 

6.  AVhat  is  the  square  root  of  390625  ?     d^u.  625. 

390625 
Multiply  by  4, 4 

1562500 
Multiply  by  4  again, 4 

6250000 
Multiply  by  4  again, 4 

25000000 
The    square   root   of  this   last   product  is   obviously 
5000;   which,  divide  by  2,  three  times,  or  by  8,  and 
we  have  625  for  the  answer,  or  square  root  of  the  origi- 
nal sum. 

7.  What  is  the  square  root  of  119025  ?    Jins.  345. 

8.  What  is  the  square  root  of  75625?       JIns.  275. 

(Art.  138.)   ABBREVIATIONS  IN  CUBE  ROOT. 

1.  What  is  the  cube  root  of  91125  ?  Ann.  45. 

Multiply  by 8 

729000 

Now,  729  being  the  cube  of  9,  the  root  of  729000  is 
90;  divide  this  by  2,  the  cube  root  of  8,  and  we  have  45, 
the  answer. 


252  ARITHMETIC. 


2.  The  contents  of  a  cubical  cellar  are  1953,125  cu- 
bic feet ;  what  is  the  length  of  one  of  its  sides  ? 

Ms.  12,5  feet. 

1953,125 
Multiply  by  8, 8 


15625,000 
Multiply  by  8  again, 8 


125,000 
The  cube  root  of  this  is  50;  divide  by  4,  because  Ave 
multiplied  by  8  twice,  and  we  have  12,5,  tlie  answer. 

When  it  is  requisite  to  multiply  several  numbers  to- 
gether, and  extract  the  cube  root  of  their  product,  try  to 
change  them  into  cube  factors,  and  extract  the  root  he- 
fore  multiplication. 

EXAMPLES. 

1.  What  is  the  side  of  a  cubical  mound  equal  to  one 
288  feet  long,  216  feet  broad,  and  48  feet  high  ? 

The  common  way  of  doing  this,  is  to  multiply  these 
numbers  together,  and  extract  the  root,  a  lengthy  opera- 
tion. But,  observe,  that  216  is  a  cube  number,  and  288 
=2X  12X  12.  and  48  =  4  X  12;  therefore,  the  whole  pro- 
duct is  216X8X  12X  12X  12.  Now,  the  cube  root  of 
216  is  6,  of  8  is  2,  and  of  12^  is  12,  and  the  product  of 
6X 2 X  12  =  144,  the  answer. 

2.  Required  the  cube  root  of  the  product  of  448X392, 
in  a  brief  manner.  Jlns.  56. 

3.  What  is  the  side  of  a  cubical  mound,  equal  to  one 
144  feet  long,  108  feet  broad,  and  24  feet  deep  ? 

.8ns.  72. 

4.  A  pile  of  wood  is  160  feet  long,  G  feet  wide,  and  9 
feet  high  ;  what  would  be  the  side  of  a  cubic  pile  con- 
taining the  same  quantity?    jj^^^  l2  3V5=20,5+feet. 

5.  What  shall  be  the  side  of  a  cubical  cistern,  to  con- 
tain 04  hogsheads?  Jins.  8,138  feet. 
Solution:  64 X  63 X 231  =04 X  7X9X3X77=64.27.539 

Cube  root  is  4X  3X='V539  =  12X8,1 38  in.,  or  8,138  ft. 


ABBREVIATIONS  IN  CUBE  ROOT.         253 


(Art.  139.)  T'fe  can  extract  the  root  of  cube  num- 
bers, by  inspection,  when  they  do  not  contain  more  than 
two  periods. 

EXAMPLES. 

Find  the  cube  root  of  1951 12.  This  number  consists 
of  two  periods;  compare  the  superior  period  with  the 
cubes  in  the  table,  and  we  find  that  195  lies  between  125 
and  2,16.  The  cube  root  of  tlie  tens,  then,  must  be  5. 
The  unit  fio-ure  of  the  given  cube  is  2,  and  no  cube  in  tiie 
table  has  2  for  its  unit  figure,  except  512,  whose  root 
is  8  ;  therefore,  58  is  the  root  required. 

The  number  912673  is  a  cube  ;  what  is  its  root? 

Ans.  97. 

Observe,  the  root  of  the  superior  period  must  be  9,  and 
the  root  of  the  unit  period  must  be  some  number  which 
will  give  3  for  its  unit  figure  when  cubed,  and  7  is  the 
only  fio-ure  that  will  ansv/er. 

The  following  numbers  are  cubes  ;  required  their  roots. 

1.  What  is  the  cube  root  of  59319  ?  Ans.  39. 

2.  What  is-  the  cube  root  of  79507  ?  Ans,  43. 

3.  What  is  the  cube  root  of  117649?         Ans.  49. 

4.  What  is  the  cube  root  of  1 10592  ?         Ans.  48. 

5.  What  is  the  cube  root  of  357911  ?         Ans.  71. 

6.  Wh.at  is  the  cube  root  of  889017  ?         Ans.  73. 

7.  What  is  the  cube  root  of  571787?         Ans.  83. 

When  a  cube  has  more  than  two  periods,  it  can  gene- 
rally be  reduced  to  two  by  dividing  by  some  one  or  more 
of  the  cube  numbers,  unless  the  root  is  a  prime  number. 

The  number  4741632  is  a  cube;  required  its  root. 
Here  we  observe,  that  the  unit  figure  is  2;  the  unit  figure 
of  the  root  must,  therefore,  be  the  root  of  512,  as  that  is 
the  only  cube  of  the  9  digits  v/hose  unit  figure  is  2.  The 
cube  root  of  512  is  8;  therefore,  8  is  the  unit  figure  in 
tlie  root,  and  the  root  is  an  even  number,  and  can  be  di- 
vided by  2 — and,  of  course,  the  cube  itself  can  be  divi- 
ded by  8,  the  cube  of  2. 

8)4741632 


592704 


25i  ARITHMETIC. 


Now,  ?,s  the  first  number  was  a  cube,  and  being  divi- 
ded by  a  cube,  tlie  number  592704  must  be  a  cube,  and, 
by  inspection,  as  previously  explained,  its  root  must  be 
84,  which,  multiplied  by  2,  gives  168,  the  root  required. 

The  number  13312053  is  a  cube ;  what  is  its  root  ? 

./??2.<?.  237. 

As  there  are  three  periods,  there  must  be  three  fiirures, 
units,  tens,  and  hundreds,  in  the  root ;  the  hundreds  must 
be  2,  the  units  must  be  7.  Let  us  then  find  the  second 
figure,  or  the  tens,  in  the  usual  ivay,  and  w^e  have  237 
for  the  root. 

Again,  divide  13312053  by  27,  and  we  have  493039 
for  another  factor.  The  root  of  this  last  number  must  be 
79,  which,  multiplied  by  3,  the  cube  root  of  27,  gives 
237,  as  before. 

The  number  18609625  is  a  cube  ;  what  is  its  root? 

As  this  cube  ends  with  5,  we  will  multiply  it  by  8 : 
18609625 
8 


148877000 
As  the  first  is  a  cube,  this  product  must  be  a  cube;  and, 
as  far  as  labor  is  concerned,  it  is  the  same  as  reduced  to 
two  periods,  and  the  root,  we  perceive  at  once,  must  be 
530,  w^hich,  divided  by  2,  gives  265  for  the  root  required. 

TO  FIND  THE  APPROXIMATE  CUBE  ROOT  OF  SURDS. 

(x\rt  140.)  The  usual  way  of  direct  extraction,  is  too 
tedious  to  be  much  practiced,  if  any  shorter  method  can 
possibly  be  obtained.  By  the  invention  of  logarithms,  a 
very  short  method  has  been  found  ;  but,  before  tliat  event, 
several  eminent  mathematicians  bestowed  much  time  and 
labor  to  obtain  short  practical  rules — and  some  of  their 
rides  are  too  ingenious  and  useful  to  be  lost,  notw^ith- 
standing  the  invention  oflrgarithras  has  nearly  superced- 
ed their  absolute  value  in  practice. 

The  following  method  is  from  Dr.  Halley's  algebraic 
formula,  but  more  commodiously  expressed ;  and  after 
knowing  the  result  of  the  analysis,  -we  can  apparently 
draw  the  same  from  mere  observations  on  numbers,  thus. 


ABBREVIATIOXS  IN  CUBE  ROOT.         255 


Let  us  take  two  cube  numbers,  say  125  and  216,  whose 
roots  are  5  and  G.  There  must  be  some  law  of  compar- 
ison between  them,  either  simple  or  complex.  We  evi- 
dently cannot  make  a  proportional  comparison  between 
them,  as  125  is  not  to  216,  as  5  to  6.  But  let  us  double 
125  and  add  216,  which  gives  466,  and  double  216  and  add 
125,  which  gives  557.  Now  466  is  to  557  as  5  to  6, 
nearly.  Here  we  have  an  approximale  proportion.  But 
to  have  a  very  near  approximation,  our  cubes  nuist  have 
a  nearer  relative  value  ;  216  is  nearly  double  of  125  :  this 
should  never  be  the  case  in  making  practical  use  of  ap- 
proximate proportions. 

To  show  that  we  can  be  more  accurate,  observe  that 
216000  and  226981  are  cubes;  their  roots  are  60  and 
61. 

Now  216000  is  not  to  226981  as  60  to  61.  But  let 
us  double  the  first  and  add  it  to  the  second,  and  double 
the  second  and  add  it  to  the  first,  and  we  shall  have 
658981  and  669962,  which  are  to  each  other  very  near- 
ly as  60  to  61. 

Or,  by  the  principles  of  proportion,  the  first  is  to  the 
difference  between  the  first  and  second,  as  is  the  third  to 
the   difference  between  the  third  and  fourth.     That  is, 

668981  :   10981   :  :  60  :   1  very  nearly. 

But  1  is  the  difference  between  the  two  roots,  and  if  the 
last  root,  or  61,  were  unknown,  this  proportion  would  give 
it  very  nearly ^  by  giving  the  difference  between  the  two 
roots. 

EXAMPLES. 

1.  Required  the  cube  root  of  66. 

The  cube  root  of  64  is  4.  Now  it  is  manifest,  that  the 
cube  root  of  66  is  a  little  more  than  4,  and  by  taking  a 
similar  proportion  to  the  preceding,  we  have 

64X2  =  128     2X66=132 
63  64 

194  :  196  :  :  4  :  to  root  of  66. 

Or,   194  :  2  ::  4  :  to  a  correction. 


256  ARITHMETIC. 

194)8,0000(0,04124 
7  76 

240 
194 

460 

388 

720 

Therefore,  the  cube  root  of  66  is  4,04124. 

2.  Required  the  cube  root  of  123. 

Suppose  it  5  ;  cube  it,  and  we  have  125. 

Now  we  perceive,  that  the  cube  of  5  being  greater  than 
123,  the  correction  for  5  must  be  subtracted. 

2X125=250      246 
Add      123       125 

As 373  :  371    :  :    5  :  root  of  123. 

Or, 373  :  2  : :  5  :  correction  for  5. 

373)10,0000(0,02681 
746 

2  540  From  5,00000 

2  238  Take  0,02681 

3020  Ans.    4.97319 

2984 

860 

From  what  precedes,  we  may  draw  the  following 

Rule.  Take  the  nearest  rational  cube  to  the  given 
number,  and  call  it  the  assumed  cube;  or,  assume  a 
root  to  the  given  number,  and  cube  it.  Double  the  as- 
sumed cube  and  add  the  number  to  it ;  also,  double  the 
number  cfnd  add  the  assumed  cube  to  it.  Take  the  dif- 
ference of  these  sums,  then  say,  Jis  double  of  the  as- 


ABBREVIATIONS  IN  CUBE  ROOT.         257 


Sinned  cube,  added  to  the  numher,  is  to  this  difference, 
so  is  the  assumed  root  to  a  correclion. 

This  correction,  added  to  or  subtracted  from  the  assum- 
ed root,  us  the  case  may  require,  will  give  the  cube  root 
very  nearly. 

By  repeating  tlie  operation  with  tlie  root  last  found  as 
an  assumed  root,  we  may  obtain  results  to  any  degree  of 
exactness  ;  one  operation,  however,  is  generally  suffi- 
cient. 

3.  What  is  the  cube  root  of  28?  £ns.  3,03658-f- 

4.  What  is  the  cube  root  of  26  ?  .fins.  2,96249-{- 

5.  What  is  the  cube  root  of  214?  Ans.  ^,m\A2-{- 

6.  Wjiat  is  the  cube  root  of  346  ?  Ans.  7,02034-1- 

The  above  being  very  near  integral  cubes — that  is,  28 
and  26  are  both  near  the  cube  number  27,  214  is  near 
216,  &c.,  all  numbers  very  near  cube  numbers  are  easy 
of  solution. 

We  now  give  other  examples,  more  distant  from  inte- 
gral cubes,  to  show  that  the  labor  must  be  more  lengthy 
and  tedious,  though  the  operation  is  the  same. 

1.  What  is  the  cube  root  of  3214?  Ans.  14,75758. 

Suppose  the  root  is  15 — its  cube  is  3375,  which,  be- 
ing greater  than  3214,  shows  that  15  is  too  great;  the 
correction  will,  therefore,  be  subtractive. 

By  the  rule,  9964  :  161  : :  15  :  0,2423,  the  correc- 
tion. 

Assumed  root, 15,0000 

Less, 2423 

Root  nearly, 14,7577 

Now  assume  14,7  for  the  root,  and  go  over  the  opera- 
tion again,  and  you  will  have  the  true  root  to  8  or  10 
places  of  decimals. 

2.  What  is  the  cube  root  of  14760213677  ? 

Ans,  2453. 

Suppose  the  root  2400,  &c.  Take  the  correction  to 
the  nearest  unit,  and  you  will  find  it  53. 


y2 


258  ARITIIMKTIC. 


3.  Wliat  is  the  cube  root  of  980922617856  ? 

.^7is.  9936. 
Suppose  the  root  to  be  10000. 

4.  What  is  the  cube  root  of  9  ?  ^ns.  2,08008. 

5.  What  is  the  cube  root  of  9^  ?  ^/Ins.  2,092+ 

6.  What  is  the  cube  root  of  41  ?  ^ns.  3,4482-1- 

7.  AVliat  is  the  cube  root  of  321  ?  .^ns.  0,847-}- 

8.  What  is  the  cube  root  of  32,1  ?  ^ns.  3,178-j- 

9.  What  is  the  cube  root  of  2,5  ?  ./ins.  1,357-1- 
10.  What  is  the  cube  root  of  7  ?  Ans.  1,9129-}- 


A  GENERAL  RULE  FOR  EXTRACTING  THE 
ROOTS  OF  ALL  POAVERS. 

(Art.  141.)  Whenever  we  propose  a  root,  we  conceive 
the  given  number  to  be  a  corresponding  power. 

If  we  propose  or  demand  the  cube  root,  we  virtually 
assert  tliat  the  given  number  is  a  third  power.  If  we  de- 
mand the  fourth  root,  the  given  number  is  understood  to 
be  a  fourth  power,  &c. 

The  following  rule  is  from  mere  inspection  of  a  gene- 
ral algebraic  formula,  first  translated  into  words  by  Dr. 
Hutton. 

Rule.  1.  Point  off  the  given  nxnnher  into  periods 
consisting  of  as  many  places  as  correspond  to  the  index 
of  the  power, 

2.  Find  the  first  figure  of  the  root  by  the  table  of 
powers,  or  by  trial:  subtract  its  power  frcm  the  left 
hand  period,  and  to  the  remainder  bring  down  the  first 
figure  in  the  next  period  for  a  dividend. 

3.  Involve  the  root  to  the  next  inferior  power  to  that 
which  is  given,  and  mulliply  it  by  the  number  denoting 
the  given  power,  for  a  divisor  ;  by  which  find  a  second 
figure  of  the  root. 

!  4.  Involve  the  2vhole  ascertained  root  to  the  given 
power,  and  sid)tract  it  from  the  first  and  second  peri- 
ods. Bring  down  the  first  figure  of  the  next  period  to 
the  remainder,  for  a  nexv  dividend ;  to  which,  fi^nd  a 
new  divisor,  as  before  ;  and  so  proceed. 


ARITHMETICAL    PROGRESSION.  259 


N.  B.   Roots  of  component  powers  may  be  obtained 
more  readily  thus : 

For  the  4th  root  take  the  square  root  of  the  square 
root. 

For  the  6th,  take  the  square  root  of  the  cube  root. 
For  the  8th,  take  the  square  root  of  the  4th  root. 
For  the  9ih,  take  the  cube  root  of  the  cube  root. 
For  the  12th,  take  the  cube  root  of  the  4th  root. 

EXAMPLES. 

1.  What  is  the  5th  root  of  916132832  ? 

9161'32832(62  ^ns, 

7776  6X6X6X6X6=7776 

6X6X6X6X5=6480  div. 

6480)13853 


916132832     62X62X62X62X62=916132832 
916132832 

2.  What  is  the  7ih  root  of  194754273881  ? 

^ns.  41. 

3.  What  is  the  9th  root  of  2  ?  ^ns.  1,080059. 


ARITHxMETICAL  PROGRESSION. 

(Art.  142.)  A  series  of  numbers,  increasing  or  de- 
creasing by  a  common  difference,  is  called  Arithmetical 
Progression. 

Thus,  3,  5,  7,  9,  11,  13,  15,  &c.,  is  an  ascending  se- 
ries, whose  common  difference  is  2  ; 

And  16,  13,  10,  7,  4,  1,  is  a  descending  series,  whose 
common  difference  is  3. 

By  inspecting  the  nature  of  a  single  series  we  may  es- 
tablish the  following  facts: 

I.  That  the  last  term  of  any  ascending  series  is  made 
up  of  successive  additions  of  the  common  difference,  to 
the  first  term. 


260  ARITHMETIC. 


II.  In  a  descending  series  of  successive  subtractions, 
one  less  times  than  the  number  of  terms. 

III.  When  tlie  number  of  terms  is  odd,  there  will  be 
a  middle  or  mean  term  ;  when  even,  there  will  be  two 
middle  terms,  or  the  middle  of  the  series  will  be  between 
two  terms. 

IV.  x\s  any  series  increases  or  decreases  regularly, 
the  value  of  the  middle  term  must  be  the  average  vahiu 
of  the  whole,  and  the  middle  term  must  be  equal  to  the 
half  sum  of  the  extreme  terms. 

From  these  observations  we  draw  the  following  rules. 

[Note.  The  terms  used  in  the  rules  show  what  must 
be  i{iven,  and  we  deem  the  technicalities  too  obvious  to 
require  explanation.] 

Rule  1.  To  find  the  last  term.  Multiply  the  com- 
mon difference'  by  the  number  of  terms  less  one,"^  and 
add  the  product  to  the  first  term,  if  an  ascending  series, 
otherwise  subtract  it. 

Rule  2.  To  find  the  sum  of  a  series.  Multiply  half 
the  sum  of  the  extreme  terms,  or  mean  term,  by  the 
number  of  terms. 

Rule  3.  To  find  the  common  difference.  Divide  the 
difference  of  tlie  extremes  by  the  number  of  terms  less 
one. 

Rule  4.  To  find  the  number  of  terms.  Divide  the 
difference  of  the  extreme  terms  by  the  common  differ- 
ence, and  add  1  to  the  quotient, 

PROMISCUOUS    EXAMPLES. 

1.  What  is  the  100th  term  of  the  series  2,  5,  8,  &c.? 
(Rule  1.)     Solution,.  .  •  .  99X3-1-2.     »^n5.  299. 

2.  The  first  term  is  5,  the  last  32,  and  the  number  of 
terms  10;  what  is  the  svnn  of  the  series?      Ans.  185. 

3.  How  many  strokes  does  the  hammer  of  a  common 
clock  strike  in  12  hours  ?  Anfi.  78. 

4.  What  debt  can  be  discharged  in  one  year,  by  week- 

*  Observe,  that  the  first  term  exists  independenily  of  the  common 
difference. 


ARITHMETICAL    TROGRESSION.  261 


ly  payments  in  arithmetical  progression,  the   first  being 
$12,  and  the  last,  or  fifty-second,  payment  $1230  ? 

^7is.  32418. 

5.  What  debt  can  be  discliarged  in  one  year,  by  weekly 
payments  in  arithmetical  progression — the  first  being  $12, 
and  the  last  $1230;  what  is  the  common  difierence  ? 

Jins.  $24. 

6.  A  man  traveling  a  journey,  went  18  miles  the  first 
day,  and  increased  his  distance  each  day  by  2  miles  ;  and 
die  last  day  went  48  miles.  How  many  days  did  he 
travel,  and  what  distance  ?  ^        C    10  days. 

'^"*'  i  528  miles. 

7.  A  houee  was  leased  for  7  years  at  $400  per  annum, 
and  the  rent  unpaid  until  the  end  of  the  lease;  how  much 
was  then  due,  simple  interest,  at  6  per  cent.  ? 

^ns.  $3304. 

N.  B.  At  the  end  of  the  1st  year,  $400  was  due;  at 
the  end  of  the  second  year,  $424  more  ;  at  the  end  of  the 
3rd  year,  $448  more,  &c. 

8.  What  will  be  the  amount  of  an  annuity  of  $50,  to 
be  paid  annually,  but  forborne  20  years ;  simple  interest, 
at  6  per  cent.?  ^7is.  $1570. 

9.  Suppose  100  apples  were  placed  in  a  right  line,  2 
yards  apart,  and  a  basket  2  yards  from  the  first ;  how  far 
would  a  boy  travel  to  gather  them  up  singly,  and  return 
with  each  separately  to  the  basket. 

^ns.  20200  yards. 

(Art.  143.)  An  arithmetical  series  may  be  represent- 
ed by  the  surface  between  two  converging  lines.  Two 
such  surfaces  put  together,  the  widest  end  of  one  against 
the  narrowest  end  of  the  other,  form  a  regular  parallelo- 
gram. 

Thus, 3     5     7     9  11 

Reversed,.  .  .  11     9     7     5     3 


14  14  14  14  14 


The  sum  of  the  two  converging  spaces  make  one  equal 
in  width,  &;c.  This  is  another  method  of  explaining 
Rule  2. 


262  ARIT."i.MtTIC, 


10.  How  many  acres  in  a  piece  of  land  80  rods  wide 
at  oae  end.  and  60  ai  the  other,  and  120  rods  long  ? 

dfu.  o'2\. 

It  may  be  observed,  that  the  natural  numbers,  1,2,  3, 
4,  5,  6.  7,  6ic.,  is  an  arithmetical  series,  whose  first  term 
is  1,  and  common  difference  1 ;  and  that  the  last  term  is 
equal  to  the  number  of  terms. 

From  this  series,  we  may  form  another  by  adding  to 
each  term  the  sum  of  cdl  the precedinsr,  and  we  shall  have 
I,  3,  6,  10,  15,  21,28,  &c. 

These  are  called  triansrular  rtvrribers,  because  ihey 
may  be  represented  by  points,  forming  equilateral  trian- 
gles, thus : 


Heace  we  perceive,  iha:  the  sum  of  the  natural  numbers, 
lo  any  degree,  expresses  the  triangular  number  of  ihe 
same  degree. 


GEOMETRICAL  PROGRESSION. 

(Art.  144.)  A  series  of  numbers,  increasing  or  de- 
creasi-ng  by  a  common  ratio,  is  called  a  Geometrical 
Prosrression. 

Thus,  2,  4,  8,  16,  32,  64,  128,  is  an  increasing  se- 
ries, whose  common  ratio  is  2  ; 

And  729,  243,  81,  27,  9,  3,  is  a  decreasing  series, 
whose  common  ratio  is  ^. 

In  arithmetical  progression,  the  terms  vary  by  con- 
stant atiditions  or  subtractions  :  here,  they  vary  by  con- 
stant multiplicHtions  or  divisions.  The  first  term  exists 
independendy  cf  the  ratio.  After  we  use  the  ratio  once 
we  have  two  terms,  using  it  twice,  we  have  3  terms,  A:c. 

AVhererer  we  slop  is  the  last  term  :  ience.  to  find  the 
last  term  we  hive  ;;..; 


GKOMF.TRICAL    PROGRKSSIOX.  263 


RuLc.  Raise  the  ratio  to  a  power  one  lesa  than  the 
number  of  terms,  and  midtiphj  it  by  the  first  term. 

To  investigate  a  rule  to  find  the  sum  of  any  series,  take 
the  following, 

3,  12,  48,  192,  768,  3072,  12288. 
Multiply  by  the  ratio  4,  and  place  the  product  one  term 
to  tiie  right, — 

12,  48,  192,  768,  3072  12288,  49152. 
Subtract  the  upper  series  from  the  lower,  and  we  have 
49152 — 3.  'J'his  difference  comprises  3  limes  the  orig- 
inal series,  because  we  multiplied  it  by  4,  and  then  sub- 
tracied  the  original,  4 — 1=3.  Therefore,  di.  ide  this  re- 
mainder by  3,  and  we  have  the  sum  of  the  series,  'i'he 
remainder,  49152 — 3,  consists  of  but  two  terms.  One  is 
the  last  term  of  the  original  series,  multiplied  by  the  ra- 
tio. The  other  term  is  the  first  term  of  the  original  se- 
ries subtracted.  We  then  divide  the  difference  by  the 
ra'io  less  one,  and  we  have  the  sum  of  the  series. 

Hence,.to  find  the  sum  of  a  series,  we  have  the  follow- 
ing 

Rule  2.  Multiply  the  last  term  by  the  ratio;  and 
from,  the  product,  subtract  the  first  term:  then  divide 
the  remainder  by  the  ratio  less  one. 

When  the  extreme  terms,  and  number  of  terms,  are 
given  to  find  the  ratio,  the  reverse  of  Rule  1  is  used. 

Rule  3.  Divide  one  extreme  by  the  otiier,  and  extract 
such  a  root  of  the  quotient  as  corresponds  to  the  num- 
ber of  terms  less  one. 

EXAMPLES. 

1.  The  first  term  of  a  series  is  4,  the  ratio  4,  and  the 
number  of  terms  9  ;  what  is  the  last  term  ? 

Ans.  262144. 

2.  The  first  term  of  a  series  is  2,  the  ratio  3,  and  the 
number  of  terms  8  ;  what  is  the  last  term,  and  what  is 
the  sum  of  the  terms  •      ^       5  ^'^'"^^  term,  ....  4374. 

'  *  I  Sum  of  the  term^,  6560. 

3.  What  is  the  sum  of  ten  terms  of  the  series  1,  f,  ^, 
&c.  Ans.  V^VV/. 


264  ARITHMETIC. 


4.  The  first  term  of  a  series  is  3,  the  last  term  12288, 
and  the  number  of  terms  8;  what  is  the  ratio,  and  what 
are  the  other  terms  ? 

Jlns.  Ratio,  4  ;  other  terms,  12,  48,  &c. 

5.  Sold  10  yards  of  velvet,  at  4  mills  for  the  first  yard, 
20  for  the  second,  100  for  the  third,  &c.;  what  did  the 
piece  cost?  ^^ns.  $^9765,624. 

6.  What  is  the  cost  of  a  coat,  with  14  buttons,  at  5 
mills  for  the  first,  15  for  the  second,  45  for  the  third,  &c. 

Ans.  $11957,42. 

7.  What  is  the  cost  of  16  yards  of  cloth,  at  .3  cents  for 
the  first  yard,  12  for  the  second,  48  for  the  third,  &lc..1 

.ans.  $42949672,95. 

(Art.  145.)  The  sinn  of  a  geometrical  series  is  found 
by  the  extremes  and  the  ratio,  independent  of  the  num- 
ber of  terms  ;  hence,  whether  the  number  of  terms  be 
many  ovfeiv,  there  is  no  variation  in  the  rule.  We  may, 
therefore,  require  the  sum  of  the  series,  6,  3,  1,  |,  f ,  &c. 
to  infinity,  provided  we  can  determine  the  value  of  the 
other  extreme.  Now,  we  see  the  terms  decrease  as  the 
series  advances;  and  the  hundredth  term,  for  example, 
would  be  exceedingly  small,  the  thousandth  too  small  to 
be  esthnated,  the  milhonth  still  less,  and  the  infinite  term 
nothing;  not,  as  some  tell  us,  "extremely  small,"  or, 
"too  little  to  be  considered,"  &:c.;  hwiab  solid  el  ij  nothing. 
Remember,  that  it  is  the  number  of  terms,  not  the  sum 
of  them,  that  is  infinite. 

Let  every  decreasing  series  be  inverted,  and  the  first 
term  called  the  last;  then  the  first  term  will  be  0,  and  the 
ratio  greater,  than  unity.  Then,  by  Rule  2,  work  the  fol- 
lowing 

EXAMPLES* 

1.  What  is  the  sum  of  the  infinite  series,  H-j+j+ 
y^^,  (fee?     Invert  the  series:  \  is  the  last  term,  and  2  the 

1X2—0 
ratio  ;  hence, =1,  the  answer. 

2.  What  is  the  sum  of  the  infinite  series  y\,  yf  7,  yi?^^, 
y\X10— 0 

&c.? =!•»  the  answer. 


GKOMETRICAL    PROGRESSION.  205 


3.  What  is  the  value  of  i-,  -^j,  j},-.,  &c.,  to  infinity  ? 

4.  What  is  the  value  of  I,  i,  i,  f ,  &c.,  to  infinity  ? 

JIns.  '2. 

5.  What  is  the  value  of  1,  |,  y^.,  &c.,  to  infinity? 

^ns.  4. 
G.  What  is  the  value  of  f,  /j,  jt-^,  &c.,  to  infinity? 

.^ns.  f . 

7.  What  is  the  value  of  ,777,  Sic,  to  infinity?  This 
may  he  expressed  by  y\,  yl^,  y^Vo'  <^c.  ^'^yi^.  ^. 

8.  What  is  the  sum  of  ,6666,  <fec.,  to  infinity  ? 

./?;2.5.  f . 

9.  What  is  the  value  of  ,232323,*&c.,  infinitely  ex- 
tended?    This  may  be  expressed  by  y^oj  toVott*  <S^c. 

10.  What  is  the  value  of  ,71333,  (fee,  to  infinity  ?  Ob- 
serve, that  the  geometrical  series  does  not  commence,  un- 
til we  pass  fJ^.  The  first  3  is  j^-~,  tlie  2d  is  y-,f^^: 
ratio  10. 

Ans.  Sum  of  the  series,  ^J-^:  whole  value,  yJ--|-_i^ 


(Art.  146.)  Geometrical  progression  is  itsedinfmd- 
ino-  (he  amount  of  anmnfies,  at  compound inf eresf,  when 
remaining  a  number  of  years  unpaid.  The  annuity  is 
the  first  term  of  a  series.  The  next  term  is  the  first  term, 
with  one  year's  interest.  This  amount  now  becomes 
principal;  and  ihis  again,  with  one  year's  interest,  is  the 
next  term,  and  so  on. 

Now  to  find  the  amount  of  any  sum  for  one  year,  we 
multiply  it  by  the  amount  of  1  dollar,  for  one  vear;  then, 
of  course,  this  multiplier  is  the  ratio  to  the  serl^«,  and  the 
sum  is  found  by  the  combination  of  Rule  1  and  Rule  2, 
of  Art.  144.  But,  to  be  less  general,  we  give  the  fol- 
lowing 

Rule.  Paise  the  ra'io  to  the  power  denoted  J.y  the 
ninnler  of  years;  multiply  that  power  by  the  annidty  ; 
fiom  Ike  product  .sidu'ract  t/ie  annuity,  and  divide  the 
rcmaiiidcr  by  the  ratio  less  1. 

Z 


266  ARITHMETIC. 


EXAMPLES. 

1.  If  an  annuity  of  125  dollars  be  forborne  4  years, 
what  will  be  its  amount,  at  compound  interest? 

Ans.  $546,81. 
(1,06)'X  125—125=32,8087  ,06)32,8087(546,81,  J]ns. 

2.  Find  the  amount  of  an  annuity  of  260  dollars,  in 
arrears  for  3  years,  at  7  per  cent,  compound  interest? 

Ans.  $835,87+ 
[NoTE.  Powers  of  the  ratio  can  be  found  in  Table  1, 
Art.  97,  as  the  numbers  in  that  table    are    nothing  but 
powers  of  interest  ratios ;  but  it  is  unnecessary  to  multi- 
ply examples.] 


ALLIGATIOx\. 


Medial  and  alternate:  a  commingling,  or  throwing 
together.  Alligation  medial,  is  finding  the  medium  or 
middle  quality,  or  value,  of  certain  amounts  of  given 
things  put  togther.  Alligation  alternate  is  finding  the 
quality  or  value  of  several  things,  corresponding  to  a  giv- 
en m.edium;  hence,  in  certain  respects,  one  is  the  reverse 
of  the  other.  Neither  of  them  are  of  much  practical 
value.  But,  as  an  improvement  to  the  mind,  and  as  a 
study  of  numbers,  they  form  a  good  lesson. 

(Art.  147.)  1st.  Alligation  medial.  From  the  defini- 
tion of  finding  a  medium,  we  recognize  the  following 

Rule.  Find  the  value  of  each  ingredient^  and  divide 
their  sum  by  the  number  of  ingredients. 

examples. 

1.  A  grocer  mixed  10  pounds  of  sugar,  at  8  cents,  with 
12  pounds  at  9  cents,  and  16  pounds  at  11  cents;  what 
was  a  pound  of  the  mixture  worth  ? 

Solution  :  Whole  cost,  364  cents  ;  divided  by  38  lbs., 
gives  9~  cts.  answer. 

2.  If  120  bushels  of  wheat  be  bought  at  80  cents  per 
bushel,  and  75  bushels  at  86  cents  a  bushel,  what  is  the 
average  cost  per  bushel  of  the  whole  purchase  ? 

Ans.  82-|-cents. 


ALLEGATION.  267 


3.  An  innkeeper  mixed  13  gallons  of  water  with  52 
gallons  of  brandy,  which  cost  him  $1,25  per  gallon: 
what  is  the  value  of  1  gallon  of  the  mixture,  and  what 
his  profit  on  the  sale  of  the  whole  at  6^  cents  per  gill  ? 

C$1  a  gallon. 
'^'^•^-  ^$65  profit. 

(Art.  148.)  Alligation  alternate.  This  requires  more 
explanation ;  and,  for  the  purpose  of  explaining,  let  us 
take  the  following  problem: 

A  grocer  has  sugars,  at  9  cents  and  16  cents  per  pound  ; 
he  wishes  to  make  a  mixture  worth  1 1  cents  ;  what  por- 
tions of  each  sort  shall  he  take  ? 

If  he  takes  1  pound  of  each,  and  sells  it  at  11  cents, 
it  is  evident  he  would  lose  5  cents  on  the  16  cent  sugar, 
and  gain  only  2  cents  on  the  9  cent  sugar.     On  that  sup- 
position, he  would  lose  more  than  he  would  gain  ;  and  to 
j  equalize  the  value,  he  must  take  more  of  the  9  cent  su- 
Igar;    as  much  more  as  9  is  nearer  ^o  11,  flum  16  is 
\  nearer  to   \l.      That  is,  take  the  reciprocal  ditference 
I  from  the  mean  price.  Ti.,,^.    n  S    ^1  ^ 

1  ^^''^'  ^^  ^16J  2 

[Note.  If  the  pupil  had  studied  Natural  Philosophy, 
I  would  compare  the  principle  of  this  process  to  balanc- 
ing weights  on  the  long  and  short  arm  of  a  lever:  the 
smaller  tlie  difference  the  greater  the  quantity ;  the  short- 
er the  arm,  the  greater  the  weight,  &c.] 

The  difference  between  9  and  11  is  2,  put  opposite  16, 
for  the  quantity;  and  the  difference  between  16  and  11 
is  5,  put  opposite  9,  for  that  quantity.  The  difference 
between  the  values  and  the  mean  value,  are  alieniafed. 
It  may  therefore  be  designated  alligation  alternate.  To 
prove  this  result  correct,  we  take  alligation  medial:  5 
pounds  at  9  cents,  cost  45  cents,  2  pounds,  at  16  cents, 
^.ost  32  ;  the  whole,  7  pounds,  cost  77  cents,  or  11  cents 
per  pound,  as  required. 

We  may  add  one  or  two  more  qualities  of  sugar  to  die 
same  question,  and  have  it  read  thus  : 

1.  A  grocer  has  sugars  at  9,  10,  14,  and  16  cents  per 


268  ARITHMETIC. 


pound,  and  wishes  to  make  a  mixture,  worth  11  cents; 
how  much  of  each  quality  shall  he  take? 

Take  a  balance  between  9  and  16,  f  9 — -.  5 

as  before;   then   between  10  and   14.        ,. J  10"^       3 
The connectinglines show  wliat  terms  "^  14J        I 

are  linked  together.    One  term  or  in-  1^16 — -^  2 

gradient  may  be  taken  to  balance  se- 
veral, when  there  is  but  one  above  or  below  the  mean 
price,  and  several  on  the  other  side. 

To  prove  that  the  above  result  is  correct,  let  us  com- 
pute the  whole  cost  and  find  the  average.  'J'hus,  5 
pounds  at  9  cents=45  cents,  3  at  10=30,  1  at  14  =  14, 
2  at  16=32.  Whole  sum,  121  cents,  divided  by  11 
pounds,  gives  11  cents  per  pound. 

2.  A  merchant  would  mix  teas  at  70,  85  and  100  cents 
per  pound,  and  sell  the  mixture  at  75  cents  ;  what  pro- 
portion of  each  must  he  take  ? 

Make  an  alternation  between  70  r70"|  "^  25-f  10 

and    100  first;    then    between    70    75  <  85  J    j     5 
and  85.     Thus,  the  answer  is,  25  (_  100--^     5 

pounds  at  70,  will  correspond  to  5 

pounds  at  100,  and  10  pounds  at  70,  will  correspound  to 
5  pounds  at  85.  Hence,  we  must  take  35  pounds  at  70, 
to  correspond  to  5  of  each  of  the  others.  Or,  we  must 
take  the  quantities  in  this  proportion,  not  necessarily 
this  quantity.  For  instance,  35  is  to  5  as  7  to  1.  Now 
if  we  take  7  pounds  at  70,  and  one  pound  each  at  85  and 
100,  the  whole  9  pounds  will  be  worth  75  cents  per 
pound.  Therefore,  after  making  a  required  mixture,  we 
can  proportion  the  different  ingredients  so  that  the  mix- 
ture shall  contain  any  required  amount  of  one  of  them. 
From  the  foregoing  we  draw  the  following 


Rule.  Set  the  several  ingredients  in  order,  one  under 
the.  other,  and  the  mean  on  one  side.  Connect  one  less 
than  the  mean  ivith  any  one  grea'er  ;  or,  if  the  case  re- 
quires if,  one  less  icith  several  greater,  or  vice  versa. 

Place  the  difference  of  each  and  the  mean  rate  against 
the  ingredient  ?oith  wliich  it  is  connected. 

If  only  07ie  difference  stand  against  any  rate,  it  will 
be  the  required  quantity  of  that  ingredient ;  but  if  there 


POSITION-.  209 


be  more  than  one,  iheir  sum  will  be  the.  quantHy  re- 
quired of  that  ingredient. 

EXAMPLES. 

1.  A  merchant  lias  spioes  at  32  cents,  10  cents  and  64 
cents  per  pound.  He  wisiies  to  mix  5  pounds  of  the  rirsl 
Willi  tile  others,  so  Uiat  tlie  compound  may  be  worth  48 
cen's.     How  mucii  of  each  must  he  use? 

./Ins.  5  pounds  of  the  second,  and  7  pounds  8  ounces 
of  the  third. 

Malve  a  mixture  first,  without  any  reg-ard  to  the  5  lbs.; 
then  if  against  the  first  stands  5  pounds,  the  problem  is 
solved,  if  not,  proportion  the  whole  so  as  to  make  the 
first  5  pounds. 

2.  A  farmer  wishes  to  mix  14  bii^hels  of  rye  worth  50 
cents  per  bushel,  with  corn  at  40  cents,  and  oats  at  30 
cents  per  bushel,  so  that  the  mixture  may  be  worth  37 
cents  per  bushel ;   what  quantities  must  he  take  of  each  ? 

^(/ns.  14  bushels  of  corn  and  32  of  oats. 

3.  A  goldsmith  has  gold  of  15,  17,  20  and  22  carats 
fine,  and  would  melt  togetlier  of  all  tliese  sorts,  so  much 
as  to  make  a  mass  of  40  ounces  18  carats  fine.  How 
much  of  each  sort  is  required? 

.^ns.  16  ounces  of  15,  8  of  17,4  of  20,  and  12  of  22. 

4.  A  man  filled  a  wine  hogshead  containing  63  gallons 
with  a  mixture  of  wine  worth  120  cents  per  gallon,  wi  h 
some  worth  160  cents,  and  water  worth  nothing.  The 
mixture  was  worth  130  cents  per  gallon  ;  how  much  of 
each  did  he  take  ? 

JIns.  9/-;  gallons  of  water,  9 /^  gallons  of  wine  at  120 
cents,  44  "n  gallons  of  wine  at  100  cents  :  or,  he  took  47  j 
gallons  of  M'ine  at  120  cents,  and  15^  gallons  at  160  cents, 
and  no  water. 


POSITION. 


(Art.  149.)  From  time  immemorial,  formal  rules  have 
been  given  in  arithmetic,  under  Position,  to  solve  a  cer- 
tain class  of  rather  complex  proportional  questions,  more 


270  AR1TH3IETI 


properly  belonging  to  algebra.  Such  rules  were  limited 
in  their  application,  and  could  cover  no  questions  involv- 
in<r  powers  and  roots,  as  powers  and  roots  are  not  pro- 
portional to  each  other. 

For  example,  16  and  64  are  square  numbers,  and  their 
roots  are  4  and  8,  or  as  1  to  2;  but  the  numbers  them- 
selves, 16  and  64,  are  to  each  other  as  1  to  4,  a  different 
relation. 

1 .  A  man  having  a  purse  of  money,  being  asked  how 
much  was  in  it,  answered,  The  square  root  of  it,  added 
to  the  half  of  it,  make  220  dollars  ;  how  much  was  in  the 
purse  ? 

It  is  evident,  that  this  question  must  be  excluded  from 
proportional  operation ;  for,  unless  Ave  first  suppose  the 
right  number,  the  result  of  the  supposition  will  not  be  to 
the  given  result  as  the  supposed  number  to  the  true  num- 
ber ;  and  when  this  proportion  fails,  supposition,  that  is, 
position  fails. 

We  prefer  to  leave  the  following  questions  without 
rules  other  than  general  analysis  and  proportion.  The 
explanations  following  some  one  or  two  will  be  sufficient. 

EXAMPLES. 

1.  One-half,  one-third  and  one-fourth  of  a  certain  num- 
ber, added  together,  make  130  ;  what  is  the  number  ? 

Jins.  120. 
We  take  the  position,  that  the  required  number  is  a 
whole.     Then, 

L-{-|-|-i.  =  ||.  Then,  by  proportion,  VI  :  130  ::  i|  : 
Jlns. 

2.  A  post  is  \  in  the  earth,  ^  in  the  water,  and  13  feet 
above  the  water  ;  what  is  the  length  of  the  post  ? 

Ans.  35  feet. 
Add  ^  and  ^  ;  not  \  and  I  of  the  number  1,  but  i  and 
■|  of  the  lohole  post.     These  parts,  added  together,  make 
II ;  the  remaining  ^1  m^^st  be  13  feet.     Then,  by  pro- 
portion. 

Or, 13     :      13     :  :     35     :     35  the  answer. 


POSITION.  271 


3.  A  and  B  have  the  same  income.  A  contracts  an 
annual  debt  amounting  to  i  of  it ;  B  lives  upon  i  of  it ; 
at  the  end  of  ten  3'ears,  B  lends  to  A  money  enough  to 
pay  off  his  debts,  and  has  160  dollars  to  spare;  what  is 
the  income  of  each  ?  Jlns.  $280. 

If  B  lives  on  j,  he  saves  t ;  out  of  this  he  pays  A's 
debts,  i.  Hence,  from  \  subtract  i-,  and  there  remains 
/j.  This,  in  10  years,  is  worth  160  dollars;  therefore, 
in  1  year  it  is  worth  16  dollars.     Now,  by  proportion, 

-"■J  :  16  ::  |i  :  the  answer ; 
Or,  .  .  2  :  16  :  :  35  :  the  answer ; 
Or,  .  .    1     :       8     :  :     35     :     280,  the  answer. 

4.  Of  the  trees  in  an  orchard,  J  are  apple-trees,  -^-^ 
pear-trees,  and  the  remainder  peach-trees — which  are  20 
more  than  |  of  the  whole  ;  what  is  the  whole  number  in 
the  orchard?  Ans.  800. 

5.  A  gendeman  bought  a  chaise,  horse,  and  harness  for 
^378 — the  horse  came  to  twice  the  price  of  the  harness, 
and  the  chaise  to  twice  the  price  of  both  the  horse  and 
harness  ;  what  did  he  give  for  each  ? 

This  problem  is  generally  given  under  position  ;  but  it 
is  really  one  in  simple  division.  Divide  the  money  into 
shares  :  it  will  take  one  share  to  purchase  the  harness,  2 
shares  to  purchase  the  horse,  &c. 

6.  A  person  being  asked  what  time  it  was,  answered, 
that  the  time  past  noon  was  \  of  the  time  to  midnight; 
what  was  it  ?  Jlns.  2  o'clock. 

Solution,  1^     :      12     :  :     \     :  Ans. 

7.  There  is  a  fish  whose  head  weighs  9  pounds,  the 
tad  weighs  as  much  as  his  head  and  half  of  his  body,  and 
the  body  weighs  as  much  as  his  head  and  tail ;  what  is 
the  whole  weight  of  the  fish  ?  Ans.  72  pounds. 

8.  A  man,  after  spending  \  of  his  money  and  \  of  the 
remainder,  had  only  10  dollars  left;  how  much  had  he 
at  first  ?  Ans.  $30. 


272  ARITHMETIC. 


PERMUTATION. 

(Art.  150.)  The  only  changes  tliat  can  be  made  with 
two  things,  as  two  lelters,  «,  /;,  are  two,  ab  and  ba.  If  we 
add  anotlier  thing  or  letter,  as  f,  we  can  place  it  on  the 
right,  on  the  left,  and  in  the  middle  of  ab,  making  3  per- 
mutations witli  each  of  the  above.  Hence,  with  2  things, 
the  numerical  expression  will  be  1X2=2,  for  3  things, 
1X2X3  =  6. 

If  we  take  another  letter,  d,  we  can  place  it  on  the 
right,  on  the  left,  and  between  each  of  the  letters,  abc, 
making  4  new  permutations,  with  each  of  the  six  ])erniu- 
tations  already  made,  or  24  in  all.  Hence,  for  four  things, 
the  expression  will  be  1X2X3X4=24.  For  any  num- 
ber of  things,  then,  we  have  the  following 

RuLK.  Take,  the,  continued  product  of  the  natural 
numbers,  1,  2,  3,  4,  &c.,  2(p  to  the  given  number  of 
things,  for  the  number  of  pennicfatims  [or  changes) 
that  can  be  made  ivilh  that  number. 

EXAMPLES. 

1.  How  many  changes  can  be  rung  on  5  bells  ? 

Ans.  120. 

2.  Seven  gentlemen  agreed  to  remain  together  as  long 
as  they  could  arrange  lliemselves  (hlTerently  each  day  at 
dmner ;  how  many  days  did  they  remain  ? 

Ans.  5040. 

3.  How  many  variations  will  ten  letters  of  the  alpha- 
bet admit  ?         '  Ans.  3556800. 

(Art.  151.)  How  many  permulations  can  be  made 
with  the  five  letters,  a,  b,  c,  d,  e,  taking  two  at  a  time  ? 

Take  a,  and  place  it  by  itself.  To  this,  each  of  the 
other  4  letters  can  be  joined,  making  4  permutations. 
'J'iie  same  can  be  done  with  each  of  the  5  letters,  making 
5X4=20  permutations  of  2  letters. 

If  we  set  apart  each  arrangement  of  2  letters,  found  as 
above,  and  connect  each  one  with  the  3  remaining  letters. 


COiMniNATlON.  273 


we  must  have  20 X  .'^,  or  5X4X3  =  60  permutations  of 
3  letters.     Hence  this 

Ri'LK.  The  permuUi'ions  ivhlch  can  he  male  of  any 
ymnil  er  cf  ihim^s,  taken  a  glve)i  number  ut  a  timc^ 
are  equal  to  the  conlinued  product  of  a  decreasing  na- 
tural series,  whose  greatest  term  is  the  ivliole  number 
of  tilings,  and  ivhone  number  of  terms,  the  number  to 
be  taken  at  a  time. 

4.  How  many  numbers  can  be  expressed  by  the  nine 
digits,  taken  four  at  a  time  ?  Ans.  3024. 

5.  How  many  words  of  five  letters  each,  may  be  made 
from  an  alphabet  of  20  letters,  supposing  that  any  num- 
ber of  consonants  may  make  a  word  ?   Ans.  7893600. 


COMBINATION. 


(Art.  152.)  Combination — is  finding  how  many  dif- 
ferent ways  a  less  number  of  things  can  be  taken  out  of 
a  greater  number  of  the  same  kind. 

How  many  combinations  of  two  letters  can  be  made 
from  tlie  letters  a,  b,  c.  d? 

Let  the  pupil  observe  that  ab  and  ha,  though  different 
permutations,  are  the  same  ccmbination,  and  the  differ- 
ent combinations  are  ab,  ac,  ad,  be,  cd,  bd  =  Q. 

No  two  are  in  any  sense  alike,  and  therefore  they  are 
in  fact  different  ccmbinadons. 

Now  tlie  number  of  permutations  of  two,  from  4 
thinjjp,  we  have  seen,  (Art.  151,)  is  4X3;  and  the  num- 
ber of  permutations  of  tico  is  1  X  2.  The  first  divided  by 
the  second,  gives  0  the  number  of  comlinations ;  and 
thus,  generally,  to  find  the  combinations  of  any  given 
number,  taken  out  of  a  given  number,  we  have  the  fol- 
lowing 

Rule.  Find  the  number  of  permutations  which  can 
be  made  from  the  proposed  set,  faking  the  given  num- 
ber of  tilings  at  a  time,  and  divide  it  by  the  number  of 
permufations  which  cctn  be  made  on  another  set,  con- 


274  ARITHMETIC. 


sisting  only  of  as  many  things  as  are  to  be  taken  at 
a  time. 

1.  How  many  combinations  of  two  letters  can  be  made 
from  24  ?  Jlns.  276. 

2.  How  many  combinations  of  3  things  can  be  select- 
ed from  20?  Ans.  1140. 

3.  How  many  syllables  of  3  letters,  can  be  found  out 
of  18  letters,  no  one  of  which  occurs  twice  in  the  same 
syllable ;  and  no  two  syllables  containing  all  the  same 
letters  ?  Ans.  816. 


MISCELLANEOUS  EXAMPLES. 

1.  The  divisor  of  a  certain  number  is  I,  the  quotient 
is  120  ;  what  is  the  dividend  ? 

2.  The  dividend  is  ,42,  the  quotient  42  ;  what  is  the 
divisor  1 

3.  How  many  men  must  be  employed  to  accomplish 
a  piece  of  work  in  15  days,  which  would  require  5  men 
21  days? 

4.  My  horse  and  saddle  are  worth  $96 ;  my  horse  is 
worth  7  times  as  much  as  my  saddle  ;  what  is  the  value 
of  each  (by  simple  division)  ? 

5.  A  can  do  a  piece  of  work  in  5  daj'S  ;  B  can  do  the 
same  in  20  days  :  how  long  will  they  be  about  it  if  they 
both  work  toijether  ?  .^ns.  4  days. 

6.  A  and  B  can  do  a  piece  of  work  in  8  days ;  A  can 
do  it  alone  in  12  days  ;  in  what  time  can  B  do  it  alone  ? 

Ans.  24  days. 

7.  A  merchant  bought  63  gallons  of  rum  for  $28,35  ; 
how  much  Avater  must  be  added  to  reduce  the  first  cost 
to  35  cents  per  gallon  ?  .^ns.  18  gallons. 

8.  $1600  was  put  at  interest  at  6  per  cent,  per  an- 
num, until  it  amounted  to  $2000  ;  what  was  the  time  ? 

Ans.  4  years  2  months. 


MISCELLANEOUS    EXAMPLES.  275 


9.  1  of  a -certain  number  exceeds  4  of  itself  by  10  ; 
what  is  the  number  ?  ^ns.  560. 

10.  Divide  a  prize  of  $10200  among  60  men,  6  sub- 
altern ofllcers,  3  lieutenants,  and  a  commander ;  giving  to 
each  subaltern  double  the  share  of  a  man,  each  lieutenant 
3  times  as  much  as  the  subaltern,  and  to  the  commander 
double  that  of  a  lieutenant ;  how  much  is  eacli  one's 
share?  ^        C  Captain  $1200. 

*^'^*-  ^Each  man  $100. 

The  three  following  problems  redeem  the  pledge  given 
in  Art.  50. 

11.  Suppose  63  gallons  jfill  a  hogshead,  42  a  tierce, 
and  36  a  l3arre] ;  what  is  the  least  quantity  of  liquid  that 
can  be  shipped  in  hogsheads,  or  tierces,  or  barrels,  just 
filling  the  vessels,  without  defect  or  redundancy  ? 

Ans.  252  gallons. 

12.  Three  men  start  at  the  same  time,  for  the  same 
point,  to  walk  round  a  circle  of  20  miles  in  circuit,  tlie 
first  going  2  miles  per  hour,  the  second  4,  and  the  third 
6  ;  how  long  before  they  come  together  again,  and  where, 
from  the  starting  point,  will  it  be  '] 

Ans.  They  will  come  together  in  10  hours,  at  the 
starting  point. 

13.  A  ferryman  has  four  boats;  one  will  carry  8  bar- 
els,  another  9,  another  15,  and  another  16  ;  what  is  the 
smallest  number  of  barrels  that  will  make  full  Ireight  for 
either  one  of  the  boats  ?  Ans.  720. 

14.  If  a  conic  pyramid  be  27  feet  high,  and  its  base  7 
feet  in  diameter;  how  many  cubic  feet  does  it  contain  ? 

Ans.  346 '3. 

15.  A  farmer  sold  17  bushels  of  barley  and  13  bush- 
els of  wheat,  for  $31,55,  the  wheat  at  35  cents  a  bushel 
more  than  the  barley  ;  what  was  the  price  of  cacli  per 
bushel  ?  ■'  /?       S  Barley  $,90. 

'^^'^•^  Wheat  $1,25. 

16.  B's  age  is  \\  the  age  of  A,  and  C's  is  2-^^  the  age 
of  both,  and  the  sum  of  their  ages  is  93  ;  what  is  the  age 
of  each  ?     [Fellowship.]  Ans.  A  12  years. 

17.  A  merchant  bought  a  pipe  of  wine  for  80  dollars, 
from  which  16  gallons  leaked  out.     What  will  he  gain 


276  ARITHMETIC. 


by  selling  the  remainder  at  12^  cents  per  pint?  and  how 
much  will  he  gain?  ^Ins.  J$30. 

18.  My  horse  and  saddle  are,  together,  worth  80  dol- 
lars ;  my  horse  is  worth  7  times  as  mucii  as  my  saddle  : 
what  is  tiie  value  of  each?  ,fins.  Saddle,  §<10. 

The  following  question  is  said  to  be  by  Sir  Isaac  New- 
ton. 

19.  If  12  oxen  eat  up  3^  acres  of  grass  in  4  weeks, 
and  21  oxen  eat  up  10  acres  in  9  weeks,  how  many  ox- 
en will  eat  up  24  acres  in  18  weeks,  the  grass  being  at 
first  equal  on  every  acre,  and  growing  equally  ? 

Jins.  36  oxen. 
'I'his  problem  requires  three  statements  in  compound 
proportion.     Thus, 

Cause.    Effect.     Cause.   Effect. 
Oxen,    .  .  12     :    '3L     :  :     21      :     [] 
Weeks,  •  •    4     :  9 

The  result  of  this  proportion  gives  13^  acres  for  21 
oxen  9  weeks.     Bui  the  question  gives  only  10  acres  ; 
therefore,  ten   acres   become   13^  by  having  5  weeks  to 
grow,  that  is,  5  weeks  longer  than  the  first  quantity. 
Cause.  Effect.     Cause.  Effect. 
Acres,.  .  •  10     :     ^     :  :     24     :     [] 
Weeks,  .  .   5  14 

The  result  of  this  proportion  gives  21  acres  of  growth, 
to  be  added  to  the  24  acres. — Total,  45  acres.     Lastly, 
Cause.  Effect.     Cause.    Effect, 
Oxen,.  .  .  12     :     3^^     :  :     []     :     45 
Weeks,  •  .    4  18 

20.  Suppose  a  meteor  so  high  in  the  heavens,  as  to  be 
visible,  at  the  same  moment,  at  Boston,  longitude  71°  3' 
W.;  at  Washington,  longitude  77°  43'  W.;  and  at  the 
Sandwich  Islands,  longitude  155°  W.:  and  suppose  the 
time  of  appearance  at  Washington  to  be  5  minutes  past 
10,  P.  M.  What  time  is  it  by  the  clocks  at  the  oilier 
places?         n       SAt  Boston,  10  h.  31  m.  40  s.,  P.M. 

•^"*-  I  At  S.  Islands,  4  h.  55  m.  52  s.,  P.  M. 

21.  What  time  is  it,  when  the  past  interval  from  noon 
is  y'y  of  the  time,  onward  to  midnight? 

Ans.  5h.  24m. 


MISCELLANEOUS  EXAMPLES.  277 


22.  A  gentleman  bought  several  gallons  of  wine  for  94 
dollars  ;  and,  after  using  7  gallons  himself,  found  that  oue- 
fourih  of  llie  remainder  was  worth  21)  dollars;  iiow  many 
gallons  were  there  at  first  ?  ^^)is.  47. 

[Solve  questions  21  and  22,  by  proportion.] 

23.  Find  three  numbers  in  arithmetical  progression, 
such,  that  the  least  shall  be  to  the  greatest,  as  5  to  9,  -.md 
ihe  sum  of  the  three,  63.  .^ns.  15,  21,  27. 

'Z4.  A  can  produce  a  certain  effect  in  3  lioiirs ;  B  in  4 
hours,  and  C  in  5  hours  ;  in  what  time  can  the  three  to- 
getiier  produce  the  same  effect?  .^ns.  ly^  hours. 

25.  A  general,  disposing  his  army  into  a  square,  found 
he  had  231  men  over  and  above;  but,  increasing  each 
side  with  one  soldier,  he  wanted  44  to  till  up  the  square; 
how  many  men  had  he]  Jlns.  19000  men. 

N.  B.  Problem  25  is  very  simple,  in  fact:  yet,  from 
some  cause  or  other,  it  is  very  troublesome  to  students — 
few  of  whom  attempt  an  unaided  solution.  2314-44  men 
are  required  to  place  round  one  side,  and  one  end,  of  a 
square  ;  how  many  men  does  such  a  square  contain  ?  is 
the  question. 

26.  A  parallelogram  is  6  rods  longer  than  it  is  wide, 
and  its  diagonal  measures  30  rods ;  required  its  area  and 
dimensions. 

.dns.  Area,  432  rods,  length  24,  breadth  18  rods. 

27.  Suppose  a  ball,  9  feet  in  diameter,  cut  down  to  a 
cube ;  what  would  be  the  length  of  a  side  ? 

^ns.  5.19+ 

28.  How  many  acres  in  a  square  field,  the  diagonal  of 
which  is  20  rods  longer  than  either  side  ? 

^ns.  14  acres  2  roods  11^-  poles. 

N.  B.  To  solve  this,  assume  any  square,  and  find  the 
difference  between  its  side  and  diagonal ;  then  make  a  pro- 
portion. 

29.  Suppose  the  hind-wheels  to  a  wagon  to  be  12  feet, 
and  the  fore-wheels  9j  feet  in  circumference;  how  far 
must  the  wagon  run,  for  the  fore-wheels  to  gain  800  rev- 
oluiicns  over  the  hind-wheels?  ^^ns.  6f|  miles. 

30.  A  hare  starts  40  yards  before  a  grayhound,  and  is 

_ 


278  ARITHMETIC. 


not  perceived  by  him  until  she  has  been  up  40  seconds  ; 
she  scuds  away  at  the  rate  of  10  miles  an  hour ;  the  dog 
makes  after  her  at  the  rate  of  18  miles  an  hour;  how 
long  will  the  hound  be  in  overtaking  the  hare,  and  what 
distance  will  he  run  ? 

^^ns.  60, f^  seconds,  and  Vv-ill  run  530  yards. 

31.  A  man  owing  the  sum  of  6480  dollars  to  several 
creditors,  orders  his  clerk  to  pay  the  first  40  dollars,  and 
the  last  500,  in  arithmetical  progression;  required  the 
number  of  creditors,  and  the  difference  between  each. 

j9ns.  24  creditors,  and  difference  $20. 

32.  Three  men.  A,  B,  and  C,  have  a  sum  of  money  to 
divide  among  themselves ;  A  is  to  have  9  dollars,  B  is  to 
have  as  much  as  A  and  \  of  C's,  and  C  is  to  have  as 
much  as  A  and  B  both ;  what  sum  is  to  be  divided,  and 
how  much  is  the  share  of  each  ? 

^        C  Whole  sum,  72  dollars. 
•^^^*-^A$9,  B$27,  C$36. 

33.  Bodies  fall  by  gravity,  about  16  feet  the  first  se- 
cond of  time,  and  the  distance  or  extent  of  fall  increases 
as  the  square  of  the  time  in  seconds  ;  how  far,  then,  will 
a  body  fall  in  7  seconds  ?         Mns.  49  X  16=784  feet. 

34.  A  body  was  observed  to  fall  during  6  seconds  ; 
how  far  did  it  fall  during  the  last  second  ? 

Ans.  176  feet. 

35.  A  body  occupied  8^  seconds  in  falling  through  a 
certain  space  ;  how  far  did  it  fall  during  the  last  half  se- 
cond ?  .^>25.  132  feet. 

36.  The  ditch  of  a  fortification  is  1000  feet  long,  9  feet 
deep,  20  feet  broad  at  bottom,  and  22  at  top ;  how  much 
water  will  fill  the  ditch?     ^ns.  1158127  gals,  nearly. 

37.  Seven  men  bought  a  grind-stone,  of  60  inches 
diameter,  each  paying  \  part  of  the  expense  ;  what  part 
of  the  diameter  must  each  grind  down  for  his  share  ? 

■Ans.  The  1st,  4,4508,  2d,  4,8400,  3d,  5,3535,  4th, 
6,0765,  5th,  7,2079,  6th,  9,3935,  7th;  22,6778  inches. 

38.  How  many  rods  less  will  it  take  to  fence  in  an 
acre,  if  it  be  laid  out  in  the  form  of  a  circle,  than  in  the 
form  of  a  square  1  ^^ns.  5|+rods. 

39.  How  many  pieces  of  cloth,  at  20,8  dollars  per 


PROMISCUOUS    EXAMPLES.  279 


piece,  are  equal  in  value  to  240  pieces,  at  12,6  dollars 
per  piece  ?  ^^ns.  145,38+pieccs. 

40.  If,  when  tlie  price  of  M'heat  is  74,6  cents  per  bush- 
el, the  penny  roll  weighs  5,2  ounces,  what  should  it  be 
per  bushel  when  the  penny  roll  weighs  3,5  ounces  ? 

Ans.  $1,108. 

41.  If  a  globe  of  stone,  9  inches  in  diameter,  weifrli 
36  pounds,  liow  much  will  another  globe  of  the  same  kind 
of  stone  weigh,  whose  diameter  is  15  inches  ? 

Ans.  166,6-4-lbs. 

42.  Mix  6  bushels  of  oats,  worth  20  cents  per  bushel, 
with  8  bushels  of  oats,  worth  25  cents  per  bushel,  to  rye, 
worth  70  cents  per  bushel,  and  wheat  worth  80  cents — 
and  sell  the  mixture  at  75  cents  per  bushel ;  how  much 
rye  and  wheat  must  there  be  in  the  mixture  1 

Ans.  Rye,  14  bush.;  wheat  160  bush. 

43.  Seven  men  not  agreeing  with  the  owner  of  a  board- 
ing house,  about  the  price  of  boarding,  offer  to  give  100 
dollars  each,  for  as  long  time  as  they  can  seat  themselves 
every  day  differently  at  dinner;  this  offer  being  accepted, 
how  long  may  they  stay  ? 

Ans.  5040  days,  or  13  years  295  days. 

44.  What  will  be  the  expense  of  paving  a  rectangular 
yard,  whose  length  is  63  feet,  and  breadth  45  feet,  in 
M'hich  there  is  laid  a  foot-path  5  feet  3  inches  broad,  run- 
ning the  whole  length,  with  broad  stones,  at  36  cents  a 
yard ;  the  rest  being  paved  with  pebbles,  at  30  cents  a 
yard  ?  Ans.  $96,70^. 

45.  In  exchanging  20^  yards  of  cloth,  of  1^  yards 
wide,  for  some  of  the  same  quality,  of  J  of  a  yard  wide, 
what  quantity  of  the  latter  makes  an  equal  barter  ? 

Ans.  34^  yards. 

46.  Suppose  a  large  wheel,  in  mill-work,  to  contain  72 
cogs,  and  a  smaller  wheel,  working  in  it,  to  contain  50 
cogs  ;  in  how  many  revolutions  of  the  greater  wheel,  will 
the  lesser  one  gain  iOO  revolutions  ?  Ans.  227  j 


1 1 


APPENDIX 


MECHANICAL    POWERS. 

The  mechanical  powers  properly  belong  to  Natural 
Philosophy,  and  not  to  Arithmetic.  The  problems  per- 
taining to  them,  in  a  numerical  point  of  view,  are  usually 
very  trifling  ;  but  the  principles  on  which  the  computa- 
tions are  based,  require  exact  thought,  and  should  not  be 
passed  over  in  a  careless  manner.  We  make  these  ob- 
servations to  guard  the  pupil  against  imbibing  the  error 
of  measuring  the  importance  of  a  thing  by  the  difficulty 
or  ease  of  numerical  computations. 

Properly  speaking,  there  are  but  two  fundameotal  me- 
chanical powers,  viz.:  the  lever  and  inclined  plane. 

From  the  lever,  we  derive  the  pulley,  and  wheel  and 
axle. 

From  the  inclined  plane,  we  derive  the  screw  and  the 
ivedo^e. 

Theoretically,  the  lever  is  an  imponderable  and  inflex- 
ible bar,  supported  on  a  fixed  point,  called  the  fulcrum. 
Let  the  line  AB  represent  it, 
resting  on  the  point  C,  call-  C 

ed  the  fulcrum,  and  weights    A j B 

that  balance  each  other  at  A 
and  B. 

'I'he  fiffure  below  represents  the  application  of  the 
power  of  the  lever  as  used  in  jjrying. 


28Q 


ARITHMETIC.  281 


In  natural  philosophy  the  term,  momentum,  means 
quantity  of  force;  and  two  bodies  of  unequal  weight  can 
have  the  same  momentum  if  their  degrees  of  motion  are 
reciprocal  to  their  weight. 

A  body  in  motion,  will  require  a  certain  resistance  to 
stop  or  overcome  its  motion  :  if  its  motion  were  twice  as 
rapid,  the  resistance  must  be  twice  as  great  to  produce 
the  same  eflect. 

If  two  bodies  move  with  the  same  velocity,  and  one 
of  them  double  the  weight  of  the  other,  it  is  evident  that 
it  would  require  double  the  resistance  to  stop  the  heavier 
body,  that  would  be  required  to  stop  the  other.  This 
amount  of  resistance  is  equal  to  the  momentum,  and  evi- 
dendy  depends  on  the  compound,  or  product  of  weight 
multiplied  by  velocity. 

When  two  bodies  balance  each  other  over  the  fulcrum 
point  of  a  lever,  the  momentums  on  each  side  of  the  ful- 
crum must  be  equal. 

Conceive  the  two  bodies,  A  and  B,  (first  figure  on  the 
preceding  page,)  to  balance  on  the  point  C,  and  give  the 
bar  a  slight  motion  round  the  fixed  point  C.  If  A  is 
twice  as  far  from  C  as  B  is  from  C,  it  is  evident  that  A 
will  move  twice  the  distance  that  B  will,  in  the  same 
time.  If  A  is  three  times  as  far  from  C  as  B  is  from  C, 
then  A  will  have  three  times  the  motion  that  B  will 
have,  &LC.  <fcc.  That  is,  the  relative  motion  depends  on 
the  relative  lengths  of  the  arms  of  the  lever. 

But  the  motion,  multiplied  by  the  weight,  gives  the 
momentum  ;  or,  which  is  relatively  the  same,  the  weights, 
multiplied  by  their  distances  from  the  fulcrum,  give  their 
relative  momenta,  which  must  be  equal  to  form  an 
equilibrium. 

Therefore,  when  two  bodies  balance  over  a  fulcrum. 
The  weight  of  one  body,  7nidtiplied  by  its  distance  from 
the  fulcrum,  is  equal  to  the  iveight  of  the  other  body 
multiplied  by  its  distance  from  the  fulcrum. 

Here  are  four  things  forming  two  equal  products, 
namely,  the  two  bodies  and  the  two  arms  of  the  lever, 
and  when  one  of  them  is  sought,  the  other  three  being 
given,  we  can  find  the  sought  term  as  a  term  in  propor- 
tion.    (Art.  75.) 

__ 


282  APPENDIX. 


1.  A  lever  16  feet  long-,  the  fulcrum  3  feet  from  one 
end,  is  in  a  stale  of  equilibrium,  liokliuff  a  weight  of  390 
pounds  ;  what  is  the  power?  Ana.  90  pounds. 

2.  'J'he  arms  of  a  lever  are,  the  one  20,  the  other  3 
feet ;  what  weight  will  a  power  of  69  pounds  balance  ? 

^dns.  400  pounds. 

3.  One  arm  of  a  lever  is  IH  feet,  the  other  6  inches, 
what  weight  will  balance  a  power  of  23  pounds  ? 

Ann.  529  pounds. 

4.  Given  a  weight  of  420  pounds,  a  power  of  40,  and 
a  lever  18  feet  long;  where  shall  we  place  the  fulcrum, 
that  the  weiglit  and  power  may  balance? 

Ans.  1^1  feet  from  one  end. 

N.  B.  This  last  problem  is  the  same  as  the  follow- 
ing: 

5.  Divide  18  into  two  such  parts,  that  one  may  be  to 
the  other,  as  420  is  to  40,  whicii  is  a  problem  in  fellow- 
ship. Indeed,  ihe  weight  caid  power  du  form  fellowship 
to  preserve  an  equilibrium. 

6.  A  lever  is  30  inches  long,  tlie  weight  at  the  two 
ends  are  12  and  20  respectively  ;  what  are  the  lengths  of 
the  arms  when  the  lever  is  in  equilibrium  ? 

An.s.  18^  and  \l\  inches. 

7.  The  difference  in  the  lengths  of  the  arms  of  a  lever 
IS  6  inches,  the  weight  on  one  arm  is  20  pounds,  and  the 
weight  on  the  other  is  30;  the  lever  is  in  equilibrium  ; 
what  is  its  length  ?  Ans.  30  inches. 


Weight  may  be  applied  at  any  point  on  the  lever,  and 
the  fulcrum  at  the  end,  as  in  the  above  figure ;  this  is 
called  a  lever  of  the  second  kind. 

'J'he  weight,  in  this  case,  rests  on  botli  ends  of  the  le- 
ver in  reciprocal  proportion  to  the  distances  on  each  side 
of  it.  If  twice  as  near  one  end  as  the  other,  double  the 
weight  will  rest  on  that  end,  &c. 


ARITHMETIC.  283 


8.  Example.  A  lever  12  feet  loii'^,  one  end  rcsliiitr 
firmly  on  a  point,  a  man  lifiinjr  with  a  force  of  75  pounds 
at  tile  other;  what  weight  would  he  be  able  to  raise  2  feet 
from  the  stationary  end  ?  ^ins.  450  pounds. 

To  analyze  this  we  observe,  75  pounds  must  rest  on 
the  man's  hand,  or  on  that  end  of  the  lever,  and  a  certain 
weight  on  the  other  end.  But  this  certain  weight  must 
bear  the  same  relation  to  75  pounds  as  2  feet  to  10  feet. 

Or,  as  2     :     10     : :     75     :     375. 

That  is,  75  pounds  on  the  man's  hand,  and  375  pounds 
on  the  fixed  point ;  therefore  the  whole  weight  is  450 
pounds. 

9.  A  man  capable  of  lifting  375  pounds,  takes  hold  at 
the  end  of  a  lever  24  feet  long  ;  what  weight  can  he  raise 
6  inches  from  the  other  end  as  a  fixed  point  ? 

Ans.  18000  pounds. 

Proposition.  When  several  weights  are  suspended  on 
one  side  of  a  lever,  at  different  distances  from  the  ful- 
crum. To  find  the  force  on  that  side,  multiply  each 
weight  by  its  distance  from  the  fulcrum,  and  add  the  sev- 
eral products  together.  We  can  then  find  what  weight 
will  balance  this  on  the  other  side  of  the  fulcrum  by  di- 
viding by  that  arm  of  the  lever. 

10.  The  two  arms  of  a  lever  are  12  and  8  inches,  re- 
spectively; 2  inches  from  the  fulcrum  on  the  short  arm  is 
suspended  20  pounds,  at  6  inches  is  suspended  30  pounds, 
and  at  8  inches  is  suspended  40  pounds  ;  what  weight  on 
the  extremity  of  the  other  arm  will  balance  these? 

Ans.  45  pounds. 

11.  A  man  weighing  150  pounds  rests  on  the  short 
arm  of  a  lever  H  feet  from  the  fulcrum,  on  the  other  side 
20  pounds  is  suspended  5  feet  from  the  fulcrum  ;  there  is 
another  weight  of  15  pounds;  where  must  it  be  placed 
on  the  lever  to  make  an  equilibrium? 

Ans.  S~  feet  tVom  the  fulcrum. 


284 


APPENDIX. 


The  wheel  and  axle  is  but  a 
continuation  of  the  lever,  or  a  se- 
ries of  levers,  filiino-  up  the  whole 
circle  round  the  fulcrum.  The 
point  or  line  on  which  the  wheel 
turns  is  the  fulcrum  of  the  lever;, 
the  semidiameter  of  the  wheel  is 
the  long  arm  of  the  lever,  the  sem- 
idiameter  of  the  axle  is  the  short 
arm  ;  but  circumferences  are  to 
one  another  as  their  respective 
semidiameters;  therefore  we  can 
compute  for  equilibrium  between 
weight  and  power,  on  the  wheel  and  axle,  the  same  as 
we  do  on  the  long  and  short  arms  of  a  lever,  considering 
the  circumference  or  diameter  of  the  wheel  the  long  arm, 
and  the  circumference  or  diameter  of  the  axle  (according 
as  which  may  be  given),  the  short  arm  of  a  lever. 

Example  1.  If  the  diameter  of  the  axle  be  6  inches, 
and  that  of  the  wheel  6  feet,  what  weight,  attached  to  the 
axle,  will  balance  16  pounds  attached  to  the  M'heel  ? 

Ans.  192  pounds. 

Solution:  As  i^     :     6     :  :     16     :     Answer. 

2.  A  large  stone,  supposed  to  weigh  about  2  tons,  or 
2000  pounds,  is  required  to  be  raised  by  a  wheel  and 
axle,  the  axle  2  feet  6  inches  in  circumference,  and  the 
circumference  of  the  wheel  18  feet;  what  power  must  be 
applied  to  the  wheel?  Ans.  211}  pounds. 

N.  B.  The  answers  to  this  question  is  on  the  suppo- 
sition that  the  machine  is  perfect,  and  experiences  no  fric- 
tion ;  but  such  can  never  be  the  case  in  practice :  the 
power  must  always  bear  a  greater  proportion  to  the  weight, 
some  say  ~  more. 

3.  The  diameter  of  a  wheel  is  12  feet,  to  which  GO 
pounds  are  applied,  and  balances  3600  on  the  axle;  what 
is  the  diameter  of  the  axle  ?  Ans.  2,4  inches. 

4.  If  the  diameter  of  an  axle  be  2  feet,  and  the  di- 
ameter of  the  wheel  20  feet,  what  power  will  balance  a 
weight  of  4000  pounds.  Ans.  400  pounds. 


ARITHMETIC. 


285 


I 


The  pulley  is  a  single  wheel,  mova- 
ble on  its  center,  with  a  cord  running 
over  it.  A  single  pulley  may  be  con- 
sidered a  perpetual  lever,  with  the  ful- 
crum in  the  center,  or  the  two  arms 
equal ;  and,  of  course,  the  weight  and 
power  must  be  equal,  to  produce  an 
equilibrio.  Hen(;e,  no  power  is  gained 
by  a  rope  running  over  a  single  wheel. 

Wlien  a  rope  runs  round  a  series  of 
wheels,  as  in  the  adjoining  figure,  call- 
ed a  system  of  pulleys,  it  is  manifest 
that  the  weight  should  be  equally  divi- 
ded among  the  ropes.  As  the  whole 
machine  is  supposed  to  be  free,  the 
wheels  can  turn,  and  the  tension  of  the 
rope  can  adjust  itself;  therefore,  there 
is  no  reason  why  one  portion  of  the 
perpendicular  line  of  rope  should  bear 
more  of  the  weight  than  another. 

Hence,  to  find  the  power,  Divide  the 
weight  by  the  number  of  lines  of  rope 
sustaining  it.  Inversely,  multiply  the 
power  by  the  number  of  ropes  to  obtain 
the  weight, 

N.  B.  Care  must  be  taken  not  to  count  the  line  of  rope 
supporting  the  power  P. 

Example  1.  A  weight  of  2240  pounds  is  sustained  by 
4  ropes,  or  which  is  the  same  thing,  by  one  rope  passing 
round  4  pulleys  ;   what  power  is  required  to  balance  it  ? 

Ans.  560  pounds. 

2.  What  weight  will  a  power  of  15  pounds  balance  by 
a  rope  running  round  6  pulleys  ?         Ans.  90  pounds. 

3.  By  the  aid  of  a  system  of  12  pulleys,  how  many 
pounds  will  a  man  sustain  who  is  capable  of  applying  a 
power  of  150  pounds?  Ans.  1800 lbs. 

4.  How  many  pounds  would  be  required  by  the  aid  of 
4  pulleys,  to  sustain  800  pounds  ?  Ans.  200  lbs. 

5.  By  the  aid  of  10  pulleys,  how  many  pounds  would 
be  required  to  balance  2000  pounds  ?       Ans.  200  lbs. 


W 


286  APPENDIX. 


The  Inclined  Plane. 
A  complete  explanation  of  the  power  or  advantage  of 
an  inclined  plane  would  lead  us  too  far;  it  is  a  })roblem 
of  the  composition  of  forces  in  mechanics.  AV^e  can  only 
assert,  that,  if  a  j)lane,  as  in  the  ad- 
joining figure,  is  twice  as  long  as  it 
is  high,  a  man  can  roll  up  on  its 
surface  twice  as  much  as  he  can 
lift.  If  the  plane  be  3  times  as  long 
us  it  is  high,  we  can  roll  up,  alontr  its  surface,  three  times 
as  much  as  we  can  lift.  If  4  times,  4  times  as  much, 
&.C.;  hence,  to  find  the  power  of  a  plane, 
Divide  its  lenglh  by  its  height, 

[Note.  This  is  on  the  supposition,  however,  that  bo- 
dies roll,  or  slide,  without  friction  :  which  is  far  from  be- 
ing true.  Nevertheless,  tliese  primary  theoretical  con- 
siderations are  essential  and  important.] 

The  amount  of  friction  depends  on  a  variety  of  cir- 
cumstances, such  as  the  smoothness  and  hardness  of  the 
plane,  the  shape  and  smoothness  of  the  moving  body, 
or  the  structure  of  the  carriage,  or  machine  on  which  the 
weight  is  carried.  Every  elevation  or  depression  along 
the  common  highway,  may  be  considered  an  inclined 
plane ;  and,  in  the  construction  of  rail-roads,  great  atten- 
tion is  paid  to  the  degree  of  elevation  of  the  inclined 
planes. 

EXAMPLES. 

1.  An  inclined  plane  is  40  feet  long,  and  5  feet  high  ; 
Avhat  is  the  advantage  of  the  plane?  .^ns.  8. 

That  is,  any  power  acting  along  the  surface  of  the 
plane,  will  move  8  times  its  weight.  Conceive  a  cord  at- 
tached to  a  body  ascending  a  plane,  and  that  cord  to  run 
over  a  pulley  at  the  vertex  of  the  plane,  and  to  it  any 
weight  attached;  this  weight  is  called  a  power.  Suppose 
this  power  to  be  20  pounds.  Now,  20X8  =  160  pounds, 
that  may  be  drawn  up  the  plane. 

2.  A  plane  is  45  feet  long,  and  3  feet  high;  w-hat  is  the 
power  of  the  plane  ?  ^8ns.  15. 

That  is,  any  power  will  draw  15  times  its  weVhtj  or 


APHKNDIX. 


'Zbl 


j\,  or  Yj  of  the  load  and  wagon  would  be  a  dead  weight 
against  the  draft. 

3.  At  a  certain  place  on  a  rail-road,  the  elevation  is  at 
the  rate  of  30  feet  to  the  mile;  what  pcu'iion  of  the  trains 
make  dead  weight  in  ascending  the  j.lane  ? 

c6'?iv.  5^i-o  =  TTTr»  "early. 
N.  B.  When  the   power  acts   horizontally,  not  along 
the  surface  of  the  p!ane,  but  along  its  hase  ;    tlun  the 
power  of  the  plane  is  found  by  dividing  the  base  by  the 
height. 

4.  Suppose,  on  a  rail-road,  there  is  an  inclined  plane 
300  rods  in  length,  and  rising  lo  the  perpendicular  height 
of  50  leet;  what  jiower  will  be  required  lo  sustain  a 
weight  of  84000  pounds  ?  dns.  14000  lbs. 

The  2ved2;e  is  formed  by  two  equal  in-  \ 
dined  planes  put  together.  Its  power  is 
governed  by  the  same  principles  as  those 
of  the  inclined  jilane.  The  power  applied 
to  the  head  or  end,  is  to  the  power  on  the 
side,  as  the  wid'h  of  the  wedge  to  its  length. 
But  we  cannot  numerically  measure  the 
force  of  a  blow  ;  therefore,  there  is  no  ob- 
ject in  giving  problems  under  this  head. 

The  screw  may  he  considered  an  inclined  plane  wound 
round  a  shaft;  one  turn  gives  us  the  length  of  the  plane, 
and  the  perpendicular  distance  between  the  threads  give 
us  the  height  of  the  plane.  Therefore,  to  find  the 
power  of  a  screw,  divide  the  circumference  by  the  dis- 
tance between  the  threads.  But,  to  move  a  screw,  a  le- 
ver-power is  applied  ;  and  on  the  principles  of  momentinn, 
the  length  of  the  turn  of  the  screw  need  not  be  consid- 
ered. By  one  turn  of  the  lever,  the  screw  is  pushed 
the  distance  between  two  consecutive  threads,  and  the 
power  is  moved  through  the  circumference  described  by 
the  lever.  But,  on  the  principle  of  momentum,  the 
power,  midliplied  by  the  circimiference  of  its  motion,  is 
equal  to  the  weight  multiplied  by  the  distance  between 
the  threads. 

Eocample  1.    If  the  threads  of  a  screw  be  2  inches 


288  APPENDIX. 


apart,  the  lever  40  inches  in  length,  and  a  power  of  60 
pounds  be  applied  to  the  end  of  tlie  lever,  what  force  is 
exerted  by  the  screw?  Ans.  7543  lbs.,  nearly. 

2.  Suppose  the  length  of  the  lever  to  be  40  inches,  the 
distance  of  the  threads  1  inch,  and  the  weight  to  be  rais- 
ed, 8000  pounds  ;  required  the  power.     Ans.  33^  lbs. 

3.  If  the  distance  between  the  threads  of  a  screw  be 
1  inch,  and  the  lever  4  feet  long,  what  weight  could  be 
sustained  by  a  power  of  350  pounds,  acting  on  the  ex- 
tremity of  the  lever?  Ans.  105556 lbs.,  nearly. 

Let  the  pupil  remember  that  these  results  suppose  no 
friction  ;  but  much  of  the  power  of  the  screw  is  destroy- 
ed by  friction,  from  |  to  j  of  the  whole. 

The  mechanical  powers  may  be  variously  combined, 
and  their  combinations  form  our  many  curious  and  useful 
machines. 

The  force  applied  to  machines  is  directly,  or  indirectly, 
referred  to  gravity,  or  pounds  weight,  drawn  up  in  a  per- 
pendicular direction.  The  agents  are  various,  such  as 
wind,  water,  animal  and  steam  power.  The  power  of  a 
man,  to  be  for  any  considerable  time  pulling  on  a  rope  to 
raise  a  weight,  is  only  equal  to  about  14  pounds ;  though 
for  a  single  exertion,  to  cease  in  a  moment,  it  is  much 
greater.  The  powers  of  a  horse,  and  to  move  at  the  rate 
of  3  miles  per  hour,  is  only  125  of  dead  weight,  much 
less  than  generally  supposed ;  and  this  is  the  average 
standard  of  horsepower.  Steam  power  is  computed,  or 
rather,  compared,  with  horse  power  ;  but  many  of  these 
forces  are  very  variable  and  indefinite,  being  so  modified 
by  circumstances,  and  the  condition  of  the  machinery. 


A 
PRACTICAL  SYSTEM 

OF 


BOOK-KEEPING, 

FOR 

MECHANICS  AND  RETAILERS, 


Book-Keepihg  is  tlie  method  of  recording  a  syste 
matic  account  of  business  transactions. 

It  is  of  two  kmds—Sinffle  and  Double  Entry.  Tli; 
former,  only,  will  be  noticed  in  this  work.  On  account 
of  the  simplicity  of  Single  Entry,  it  is,  perhaps,  the  best 
which  can  be  recommended  to  farmers,  mechanics,  and 
retailers.  It  consists  of  two  principal  books — the  Day 
Booh,  or  Waste  Book,  and  the  Leger,  and  one  auxiliary 
book,  the  Cask  Book. 

THE  DAY  BOOK. 

This  book  is  ruled  Avith  a  column  on  the  left  hand 
for  the  date,  and  three  columns  on  the  right,  the  first, 
for  the  folio  or  page  of  the  Leger,  to  which  the  account 
is  transferred  ;  and  the  last  two  for  dollars  and  cents. 

This  book  exhibits  a  minute  history  of  business  trans- 
actions in  the  order  of  time  in  which  they  occur,  Avith 
every  circumstance  necessary  to  render  the  transaction 
plain  and  intelligible. 


Cincinnati,    1852. 


Jan. 


11 


11 


14 


14 


15 


David  Judkins,  Dr. 

To  10  lbs.  coffee  at  17  cts.  $1  70 

'*   25lbs.  suo-ar,atlOcts.  2  50 


Timothy  ^Y.  Coolidge,      Dr. 
To  1  bl.  sugar,  weighing 
135lbs.  neat,  atSActs.^ll  47^- 
*'  1  bag  coffee,  98  lbs.  at 
15  cents,  14  70 


George  H.  Eaton,  Dr- 

To  1  bl.  flour,  83  87^ 

"   1  lb.  Y.H.  Tea,  1  12^ 

*'   1  keg  lard,  neat  weight 

60  lbs,,  at  61  cts.  3  75 


David  Judkins, 
By  cash  on  account. 


Cr. 


James  Wilson,  Dr. 

To  3  weeks  boarding,  at  2  dol- 
lars per  week. 


Hiram  Ames,  Dr. 

To  1 2  yards  broadcloth  at  6  dol- 
lars "per  yard,  87^  00 
*'  30  yds.  muslin  at  14  cts.  4  20 
^  Cr. 
By  an  order  on  J.  Jones  for 

Groceries,  851 00 

*'  Cash,  20  00 


Timothy  W.  Coolidge,       Cr. 
By  a  bill  of  carpenter  work. 


James  Wilson, 


Dr. 

83  25 
1  keg  lOd.  nails,  weight 
1 1 1  lbs.  at  8  cts.  8  88 


To  1  bl.  vinegar 


cts. 


20 


26 


76 


71 


25 


12 


75 


00 


00 


20 


00 


00 


13 


Cincinnati,   1G52. 


Jan.     19 


21 


Feb. 


25 


29 


30 


31 


George  Hamilton,  Cr. 

By  1  set  of  Fancy  chairs, 


George  H,  Eaton,  Cr. 

By  cash,  to  balance  account, 


Kobert  Young,  Dr. 

To  cash  on  account,  $3  00 

«'  10  lbs.  N.  0.  sugar,  at 

10  cents,  100 

"  12  lbs.  coffee  at  16^  cts.  2  00 
-  J-ib.Y.H.  Tea,  62^ 


Jackson  Moore,  Dr- 

To  21  lbs.  Ham,  at  lOcts.  82  10 
1  box  soap,  30  lbs.  at 
5  cents,  1  50 


Cr. 


David  Judkins,  Di 

To  12  bis.  apples  at  75  cts. 

By  47  bu.  corn,  at  20  cents, 

James  Wilson,  Cr. 

By  cash  on  account. 


10 


Thomas  Hilton,  Dr. 

To  cash,  $5  00 

**  1  lb.  Y.  H.  Tea,  1  06 

*'  16  lbs.  Rice,  at  6^  cts.     100 

Cr. 
By  13  days  labor,  at  87|  cts. 

George  Hamilton,  Dr- 

To  1  canister  Imperal  Tea, 
14  lbs.  at -^1  87iperlb. 


•,?    ds. 
25  00 


75 


Hiram  Ames,  Cr. 

1 3 '  By  order  on  E.  Disney  for  goods, 


621 

60 

00 

40 

00 


7 

06 

11 

37i 

26 

25 

10 

00 

Cincinnati,    1852. 


Feb. 

1 
15 

21 

27 
1  6 

6 

8 
8 

16 

Robert  Young,                Dr. 
To  1  box  sperm  candles, 

25  lbs,  at  30  cts.  perlb.  ^7  50 
'*  2  bu.  dried  fruit,  at  ^125 
per  bushel,                         2  50 

(< 

James  Vv^lson,                Dr. 
To  10  gals,  molasses,  at  40 

cents,                               i4  00 
'*  4  lbs.  Old  Hyson  Tea, 

at  93  cents,                       3  72 

a 

Robert  Young,                Cr. 
By  12  cords  of  wood  at    ^2  25 

Marc! 

Ames  &  Smith,              Dr- 
To  18  lbs.  sole  leather,  at 

25  cents,                          S4  50 
'*  1  side  upper  leather,       2  75 
**  3calf  skms,  ^1  25,         3  75 

tt 

Hiram  Ames,                  Dr. 
To  500  feet  white  pine  boards 

at  812  50  per  M.            ^6  25 
**  25bu. potatoes,  at50cts.  12  50 
-1  ton  of  Hay,                 10  00 

it 

David  Judkins,               Dr. 
To  200  lbs.  tlour,  at          81  75 

tt 

Thomas  Hilton,              Dr. 
To  cash  paid  his  order  to. 
Wilham  Coohdge, 

ti 

George  Hamilton,          Dr. 
To  1  copy  Whelpley's  Com- 

pend,                                81  25 
*'  1  ream  letter-paper,        4  50 
'*  1  doz.  Spelling  books,      1  00 

10 


27 


11 


28 


Cincinnati,    1852. 


Mar. 

20 

'i 

(( 

<c 

23 

Theodore  P.  Letton,      Dr. 

To  sliarpenino-  his  plough,^!  t)0 

**  Shoeing  his  horse,  1  62^- 

"  Repairing  chain,  25 


Timothy  W.  Coolidge,     Dr 
To  2  qrs.  tuition  of  hiniself  at 
evening  school,  at  83  per  qr. 


25 


26 


Thomas  Hilton,  Cr. 

By  the  hire  of  his  horse  10  ds. 
at  621-  cts.  per  day, 


Ames  &  Smith,  Cr. 

Byl  hhd.  sugar,  weight  13171b. 
neat,  at  7|-  cents. 


T.  P.  Letton,  Dr. 

To  1  ream  wrapping  paper, 

81  C2^- 
"  1  beaver  hat,  5  00 

"  set  silver  tea-spoons,       6  00 


30 


31 


31 


Henry  C.  Sanxay,  Cr. 

By  my  order  on  him  in  favor  of 
Jno.  Torrence  for  stationary. 


Ames  &  Smith,  Dr- 

To  2000  ft.  clear  pine  boards,  at 

820  per  M.  ^40  00 

'*  500  common  do.  at  $8,    4  00 

*'  5000  shingles,  at  82 25  11  25 

"  Cashtobalanceacc't,     32  52 


Jackson  Moore, 
By  painting  my  house. 


Cr. 


98 


12 


12 


ds. 


87J 


00 

25 
77i 


62i 


871. 


87 
25 


7/ 


00 


George  Hamilton,  Cr. 

By  an  order  on  J.  Hulse,  85  00 

cash,  6  50 


12   50 


Cincinnati,    1852, 


March  31 


31 


31 


31 


Thomas  Hilton,  Dr. 

To  4  bii.  wheat,  at  ^1  25,  $5  00 

"  1  bl.  mess  pork,  9  00 

*'  2  bu.  salt  at  50  cts.'         1  00 

"  8  lbs.  brown  suo-ar,  1 1  cts.  88 


George  Hamilton.  Dr. 

To  12cedar  posts,  at  25c.  ^3  00 

'*  1  plough,  9  37| 

''  1  scythe,  1  62^ 


Jackson  Moore,  Dr. 

To  repairing  his  wagon  and 
plough, 


Thomas  Hilton,  Cr. 

By  1  pair  shoes,  ^1  50 

"    1  mahogany  table,        12  50 


15 


14 


14 


cts. 


00 


00 


00 


THE    LEGER. 

This  book  is  used  to  collect  the  scattered  accounts  of  the  Day 
Book,  and  to  arrange  all  that  relates  to  each  individual,  into  one 
separate  statement.  The  business  of  transferring  the  accounts 
from  the  Day  Book  to  the  Leger,  is  called  posting. 

The  Leger  is  ruled  with  a  double  line  in  the  middle  of  the  page, 
to  separate  the  debts  from  the  credits.  Each  side  has  two  col- 
umns for  dollars  and  cents,  one  for  the  page  of  the  Day  Book  from 
which  the  particular  item  is  brought,  and  a  column  for  the  date. 

When  an  account  is  posted,  the  page  of  the  Leger  on  which  this 
account  is  kept,  is  written  in  the  column  for  that  purpose  in  the 
Day  Book,  and  also  the  page  of  the  Day  Book  from  whicli  the  ac- 
count was  posted,  is  written  in  the  2d  column  of  the  Leger. 

In  posting,  begin  with  the  first  account  in  the  Day  Book,  which 
you  will  perceive  is  the  name  of  David  Judkins.  Enter  his  name 
in  the  first  page  of  the  Leger,  in  a  large,  fair  hand,  with  Dr.  on 
the  left  and  Cr.  on  tlie  right.— As  there  are  several  articles  charg- 
ed to  D.  Judkins  on  the  4th  of  January,  instead  of  specifying 
each  article  in  the  Leger,  we  merely  say.  For  Sundries,  and  en- 
ter the  amount  in  the  proper  columns — see  Leger,  page  1. 

The  Leger  has  an  index  or  alphabet,  in  wliich  the  names  of 
persons  are  arranged  under  their  initial  letters,  with  the  page  in 
the  Leger  where  the  account  may  be  found. 


ALPHABET    TO  THE  LEGER. 


A 
Ames,  Hiram  — 
Ames  &  Smith . . 

.2 
.'.4 

I  &  J 
Judkins,  David. 

.1 

R 

B 
Balance 

.5 

K 

S 
Sanxay,  H.  C. 

....5 

C 
Coolidge,  T.  "W.. 

.2 

Lctton 

L 
T.  P.... 

.5 

T 

D 

Moore, 

M 

Jackson. 

.4 

U 

E 

Eaton,  George  H. 

..2 

N 

y 

F 

0 

Wilson,  Jas. , . 

....3 

G 

P 

X 

H 
Hilton,  Thomas,. 
Hamilton,  Geo.... 

.3 
.3 

Q 

Y 

Young,  Robt.. 

...4 

Z 

Dr. 


David  Judkins, 


Cr. 


1852 
Jan.  4 
*'  30 
Mar.  7 


Ap.  1 


To  sundries 
''    Apples 
*'    Flour 


To  balance 
of  account 
brot.  down, 


4;20 
900 
3,50 

16  70 


4  30 


Jan.  8 

"   30 


By  cash 
''  corn 
"     Bal. 


00 
40 
30 

70 


IfoTE. — 'The  Dr.  on  the  left  hand  side  of  the  page  signifies 
debtor,  and  that  the  sums  entered  on  that  side  of  the  page,  are 
those  for  articles  sold  to  others,  and  for  which  theij  owe  you.  The 
Cr.  on  the  right  hand  side  signifies  credit,  or  that  tlie  sums  en- 
tered on  that  side  of  the  page  are  for  articles  received  of  the  per- 
son under  vrhose  account  they  stand,  and  forwhicli  you  otcehim. 


Dr. 


Timothy  W.  Coolidge, 


1852 
Jan.  4 
Ma.20 


To  sundries 
''    Tuition 


Apl.  1  To   Balance 


2,26 
5    6 

32 

7 

in 
00 

111 

1  Ja.  14 

Apl.4 

By  work 
'*  Balance 


2  25  00 
7  171 


32 


17] 


Note. — The  Dr.  side  of  this  account  shows  the  amount  of  ar- 
ticles Mr.  Coolidge  has  received  of  me,  and  the  Cr.  side  shows 
what  I  have  received  of  him.  It  appears  that  the  total  amount 
of  my  account  against  him  is  $32  17)^,  from  which  deduct  the 
$25  00  (which  stands  to  his  credit  on  the  right  of  the  account) 
and  there  will  remain  a  balance  of  $7  17}^  due  me. 

George  H.  Eaton. 


Jan.  6   To  sundries  2     8  75  Jan21    By   cash  3 


75 


JS'oTE.— This  account  presents  equal  suins  on  both  sides;  hence 
it  is  evident  that  I  owe  G.  H.  Eaton  nothing,  and  that  he  owes 
me  nothing.     The  account  is  fully  closed. 


Hiram  Ames. 


Janll 
Mar.6 


Api.  1 


To   sund. 
*'    corn 


To  Bal. 


2 

76! 

20 

Jan.  11 

4 

28j 

75 

Feb. 13 

1 
104 

95 

Apl.    1 

— - 

= 

23 

1 

1 

95 

By  sund.|2(  71100 


Order3 
Bal. 


lOjOO 
2395 

104195 


Dr.                  Thomas  Hilton,                    Or.         3 

1852 

1     , 

1 

1                    '    1 

Feb.  3 

To   simd.  3     7 

06  i 

Feb.  3  By  labor, 

3 

11 

37^ 

Mar.  8 

'*    Cash. 

4    18 

75 

Ma  23 

"■  hire  of 

'*  31 

*'      do. 

6    15 

88 

horse, 

5 

6 

25 

"  31 

"  Sund. 

6 

14 

00 

Apl.l 

-      Bal. 

1 

10 

06 1- 

41 

69 

41 

69 

Apl.l 

To  Bat. 

|io 

061 

Jca7ies   Wilson. 

Jan  11    Toboard- 

Jan31 

By  cash 

3 

|15 

00 

ing, 

2 

6 

00 

Apl.l 

*'      Bal. 

'lO 

85 

"   15 

*'  simds. 

2 

12 

13 

Feb21 

''  do. 

4 

7 
25 

72 
85 

S 

85 

Apl.l 

To    bal- 
ance bro't 
down, 

10 

85 

ISToTE. — When  au  account  is  settled  only,  and  wot  fully  paid,  as 

in  the  above,  and  several  preceding  accounts,  the  balance,  ^vvlieth- 

er  it  be  in  your  favor  or  against  you,  is  brought  down  and  placed 

distinctly  by  itself,  and  serves  for  the  beginning  of  a  new   ac- 

count, as  you  perceive  has  been  done  in  the  above  example,  the 

balance  being  $10  8.5. 

George  Hamilton. 

Feb  10 

To     Tea, 

3  |26 

25 

Janl9 

Bychairs, 

3 

25 

00 

Ma.  15 

-    Sund. 

4 

6 

75 

Ma.31 

"  Simds. 

5 

n 

50 

''    31 

-      do. 

G 

14 

47 

00 
00 

Apl.  1 

Balance, 

K 
47 

50 
00 

Apl.l 

To     Bal. 

10 

50 

Br. 


Robert    Young, 


Cr. 


1852 

Jan27 

Feb  15 

Apl.  1 


To  Sund- 
ries, 
"     do. 
'«      Bol 


62i 
00 ' 
371 


00 


Feb27 


By  wood. 


Apl.l 


By  Bed. 


00 


00 

37J 


ISToTE. — 111  tlie  above  account  the  difference  between  the  Dr. 
and  Or.  side  is  $10  STi^,  by  whicli  I  perceive  that  the  balance 
against  vie,  in  favor  of  Robert  Young,  is  $10  37i,<. 

Jackson  Moore. 


Jan29 

Ma.31 

Apl.  1 


To     Sun- 
dries, 
"  Sunds. 
-      Bed. 

1 
3 
6! 

I 

11 

60 
00 

40 

00 

Ma.31 

Apl.  1 

By  paint- 
ino\ 


By  bal- 
ance bro't 
down, 


2100 


21 


11 


00 


40 


Ames  &  Smith. 


Mar.  6 
''  30 


To  Sund. 
''    do. 


4  11 

87 

98 


00 

77 


77 


Ma. 25  1  By  sugar. 


98  77 


9877 


Note — This  account,  like  the  one  on  the  second  page,  is  fully 
closed,  the  amount  on  the  Dr.  side  being  just  equal  to  that  on 
the  Or. 


T.  F.  Letton, 


Cr 


1852 

Ma.20 
*'  26 


Apl.  1 


I  I 
To  sunds.  5 

"    do.       5 


To     Bed. 


287i 

12621 

is!  50 
15  50 


Apl.l 


By     Bal 


50 


50 


Note. — In  the  above  account  there  is  no  sum  on  the  Cr.  side, 
and  tlie  inference  is,  that  T.  P.  Letton  owes  me  $15  50  for  sun- 
dry articles  expressed  in  detail,  in  the  Day  Book,  page  5. 


Henri/  C.   Sanxay. 

Apl.l 

To    Bal. 

12871 

Ma.28 

By    my 

order, 

5 

1 

12  071 

12  871 

12 

871 

1      ; 

1      1 

Apl.l 

By  Bal. 

. 

12 

^^ 

Note. — In  this  account  it  will  be  perceived  that  there  is  no 
amount  charged  to  H.  C.  Sanx»y,  on  the  Dr.  side,  I  owe  him 
$12  87>^. 


J)r. 


Balance. 


Cr. 


Apl.l 


To  D.  Judkins, 
T.  Yr.  Coolidge, 
H.  Ames, 
T.Hilton, 
J.  Wilson, 
G.  Hamilton, 
T.  P.  Letton, 


4  30 

7,17 

•2;  23:95 


1006 
1085 
10.50 
15  50 


82.33  I 


Apl.l 


By  R.  Young, 
"  J.  Moore, 
"  H.Sanxay, 


34 .54 


Note. — This  account  exhibits  the  exact  state  of  your  books. 
It  is  made  from  the  preceding  accounts  in  the  Leger.  The  Dr. 
side  is  an  exhibit  of  the  amounts  due  to  you  by  others,  and  the 
Cr.  side  the  amounts  due  hy  you  to  others.  It  is  not  strictly  ne- 
cessary that  this  account  should  be  introduced  in  the  Leger  in 
single  entry  :  it  will  be  found  convenient,  however,  to  balance  the 


.1 


The    Cash  Boole. 


book  at  stated  intervals,  and  transfer  tlie  balances  to  the  new 
accounts  below,  as  in  the  preceding  Leg-er,  and  when  that  is 
done,  a  balance  account  like  the  above,  will  be  found  convenient, 
as  presenting,  at  one  view,  the  exact  state  of  your  Leger. 

FORM    OF    A   BILL   FROM    THE    PRECEDIXG. 

Mr.  David  Judkins, 
1852  To  Edward  Thomson,  Br. 

To  10  lbs.  coffee  at  17  cts $1  70 

25  lbs.  suo-ar  at  10  cts 2  50 


Jan. 

4 

(C 

30 

Mar 

8 

Jan 

8 

30 

12  bbls.  apples  at  75  cts. 
200  lbs.  flour  at  $175,.. 


Cr. 
By  Cash, P  00 

47  bushels  corn  at  20  cts 9  40 


Errors  excepted.  Balance  due,'   4.30 

Rec'd  Payment  in  full,  EDWARD  THOMSOJS".  — '— 


4  20 
9  00 
3  50 


16 


12 


70 


40 


THE  CASH  BOOK. 
The  Cash  Book  is  used  to  record  the  daily  receipts  and  pay- 
ments of  money.  It  is  ruled  nearly  the  same  as  the  Leger  ;  the 
Dr.  side  exhibits  the  amonntof  money  received,  and  the  Cr.  side, 
the  amount  paid  mit.  Subtraat  the  sum  of  the  Cr.  from  that  of 
the  Dr.  and  the  balance  will  always  be  equal  to  the  amount  of 
cash  on  hand. 

FORM   OF   A   CASH  BOOK. 


Dr. 


Cash. 


a-. 


1852 

Jan.  1 

"     1 

"    1 

"    1 


To  cash  on  hand. 
Cash  of  J.  Young, 
H. Sanxay, 
'  D.Juclkins, 


I     ',1852  i 
73  81  j 'Jan.  1  By  house  rent  pd. 
16  40  ' 


25  00! 
lOiOO 


T.  P.  Lctton, 
iPd.  note  R.  Hand, 
1  Family  expenses, 
1  By  cash  on  hand, 


!  125  21 


Jan.2iTo  cash  on  hand,  1 
"  2CashofTCoolidge 
•'    2  Cash  found. 


Jan. 3  To  cash  on  hand. 


52  84  By  cash  pd.  Ames 

2316  Jan.  2     &  Smith, 

29  00  !  "    2  By  cash  on  hand, 


105,00; 

"85  001 


18 

00 

50 

00 

4 

37 

52 

34 

125 

21 

= 

= 

!  20 

00 

!  S5 

00 

— 

, — 

105 

00 

^^ 

^^^ 

i 

p 


VB  35835 


JACOB  EENSJl 

ir 

iiei  irrr;}»  mi'eei,  ViticimiuU,  Ohio, 

}3ubli£ii!cr  of  Hobinson  s  i^lathcuialkal  Scriee. 


i 


do.  I7aii;\;.i  ;  ;;'losO; 

do.         (^eon'eii_ 

do.         .vstiYx;.  chooi  FditJL 

ISi""l::  •  ■   of   AlAiKU 

W"Ai--;cu  ;;  r^nooi  Dictioriu; 

N'^"  "^  n"=!  Eieiner.t?  ofScicr 

1'he  •  ^  rraoiiiu.  a  collectl'^ri  oi  i  ,;:<y  o ^ ;  . 

School  'ros*.?.mc-uts,  ll'ino. 


ALL  THU  L£ii:.  IT. 


Blank  Ecoir.CcpyBockr 
position   Bcl'.>-.  Sera     "^ 
i   Cap,  Comr'i- 
I    Drawing  i'oi      .  >     .^    ^ 
i       Blanks    f  ai!  i:^.e  varlcn- 


■3  voices,  38       llj 


■ ''  -  TE 


.u-iCoiu'  :j 


.JJ! 


